An important general point is this: when you square both sides of an equation, your new equation can have more solutions than the one you started with. Ex: x = 2 has one solution, x^2 = 4 has two solutions. When you solve an equation after having squared both sides of the original, your new equation may have one or more "false roots". Thus, it is necessary to take your bottom line "possible solutions" back to the original to see which ones don't work.
Yes, thank you. A teacher pronouncing that "only the principal square root matters in this case" tells the student nothing useful whatsoever. I already knew the answer, and this video almost sucked that knowledge out of my brain.
That's true, but it's not really relevant here. Going through a whole process of solving the equation (which involves squaring both sides) only to find that the solution doesn't work when you plug it back in to the original equation is a complete waste of time (not to mention a failure of understanding) when we can see by inspection that the equation is impossible.
This is true whenever you apply a non-injective function to the beginning equality. For example, x=2 If we apply f(x)=x+1/x to both sides, we get x+1/x= 5/2, which has 2 solutions. Mainly, x_1= 2 and x_2 = ½
Thanks for that. I was very frustrated after sitting through the lesson, because he didn't explain clearly (at least to me) what had gone wrong. I thought, if you do the same thing to both sides of the equation, the resulting equation is guaranteed to be correct. If you follow all the rules and get the wrong answer.... wha??? That was the feeling I was left with. But your general point cleared that up for me. It was just what I needed. The teacher made it sound like the problem was the definition of a symbol, which seems like just a matter of convention, an arbitrary rule, not a mathematical truth. That didn't explain how you can get the wrong answer by following a rule that should guarantee the preservation of truth. Your example showed that the problem of extra roots didn't depend on the radical symbol. It was a consequence of the squaring operation.
This is a really important thing to understand. I don't know the exact percent of students who don't get it, but from decades of tutoring, I know it's high. From the comments there are some adults who are also confused by it. (Which is okay.) If you ask the solution of x² = 4, it's x ∈ {-2, 2}. But if you ask what is the solution of √4 = x, then the solution is ONLY x=2. The square root symbol means ONLY the positive square root, or what they call the "principal square root." Start with: x² = 4 Take square root of both sides: ±√x² = ±√4 Simplifying: x = ±2, or x ∈ {-2, 2} This is why in the quadratic equation, they use the ± symbol - because the square root symbol by itself refers to ONLY the positive (principal) square root and since the equation needs to account for BOTH square roots, they have to include the plus or minus sign.
The problem here is that there is a distinction between mathematics meant to have single answers and mathematics used to solve actual real world problems.
Except for 'checking the answer with the original problem', you basically explained the whole 17+ minutes video in less than a minute of reading. Thank you, well done.
In this case it would be informative to move the - to the numerical side of the equation. It would yield a sq root equalling a negative number which is an obvious thing to look for.
@@ozviking8052 That's not a solution. That gives 2m+2 = 25i. But √(25i) does not equal -5. I think you have the relationship backwards. It looks like you're thinking √i = -1, but that's not correct. The relationship is √(-1) = i.
3-√(2m+2)=8 √(2m+2)=-5 √x is the primary functional branch of the inverse relation of the function f(x)=x², with branch cuts such that f₁⁻¹(x)≥0 and f₂⁻¹(x)
The radical has a negative of an imaginary operator on it, Heaviside's j=±1 rotating signs with each operation that squares to -1. When we divide both sides by that, we rotate it into another operator that does not flip its signs and squares to +1. It eliminates the extraneous solution. 3-±√2m+2=8 → -j√2m+2=5 → √2m+2=5/-j rationalizes to 5j'=√2m+2 → 25=2m+2
@@peppermann wikipedia also states that the radical denotes the principal square root, also called "THE square root" with a definite article, which is positive. The radical symbol is representing the square root FUNCTION, which takes the real numbers exclusively to the non-negative complex axes. A function maps every element of its domain to exactly one element in its codomain. In general "a root" is a solution to a polynomial equation in one variable. An "n'th root of k" is a solution to the polynomial xⁿ-k=0. The roots of x²-4=0 are {2,-2}, but √4 is only 2.
@@ProactiveYellow using a definite article to imply something significant. At first glance that seems to be practice that should be disavowed, but thinking back, a lot of statements in math also make quite heavy usages of grammar rules. But "THE" here really is a thing that's easy to miss.
I got "no solution" a slightly different way. I don't consider "-(sqrt(2m+2)) to be fully isolated, since there's a -1 being multiplied by (sqrt(2m+2)) there. So I divided through by the -1 to get sqrt(2m+2)=-5. At which point you already know there is no solution, since a sqrt can't give a negative answer.
Actually a sqrt can give a negative number. Without complex numbers you can't take the sqrt of negative number. - 5 * - 5 = 25 and 5 * 5 = 25 so the sqrt of 25 is +/- 5.
@@alexaneals8194 The problem here is you both are talking about different sqrt functions. You need to agree first which sqrt to use. One of them can give values in the form of negative real numbers and the other one is not allowed to do that.
@@evgtro8727 It's actually not which sqrt to use. It's the fact that they have introduced the concept of principal sqrt after I attended school. I took Algebra I in the 8th grade in 1978-1979 and Algebra II the following year 1979-1980. At that time there was no principal sqrt. The sqrt equation was assumed to include both positive and negative values. So, the roots to 25 would be +5 and -5. Even the math and engineering handbook that I have from before 2000 does not mention the principal sqrt. So, the concept is new to me. It makes sense since most people use calculators in math class and the calculator like the programming languages treat the square root as a function and not as an equation (which can return more that one value for a given input).
What promotes the era is when teachers self contradict when they write on the board something like rad(4) = +/-2. That is never acceptable. In this series of math lessons I’ve seen that done frequently. You cannot ever justify writing that a radical is equal to plus or minus anything. There is no justification for claiming that. Doing that is what creates the problems with radicals.
You are wrong. Two negatives cancel so it is acceptable. -(2)^2 =4. The problem is the new math and it's insane bias against complex numbers at the prealgebra stage when it is required for real world applications using complex analysis.
@@josephmalone253wtf? Who tought you math? First: -(2)^2 = -4 . Put the "-" within the parenteses if you want it to be squared, too...... Second: the radical sign is DEFINED to be the principal root. Always has been. Therefore √4 = +2 and nothing else. If you want to solve x^2 = 4, you HAVE to write x = +- √4 what can be simplified to +-2. Everything else is incorrect notation. And if you knew ANY math at all, you would KNOW that even complex numbers do not solve this. The radical sign is defined the exact same way for complex numbers. Don't need to trust me - try Wolfram Alpha. It'll tell you that there is no solution.
@@wernerviehhauser94I think the problem relies on people not realizing the context of the operation. Yes the radical is defined as the principal root, which is positive but then people get confused in operations like solving for x^2 because then you have 2 roots. If it's just a number with no other context but no relation to anything else then the principal root is the answer. But in this scenario I think the right approach was not to give in the urge to solve for m and just analyze the operations. 3 - (any positive number) =/= 8
@@marilynman I also assume that the culprit here is, apart from "New Math" and the educational system, the use of "root" for two distinct things: roots of a number and solutions of polynomial equations. Here in Germany, you don't find many people making this mistake since we have "Wurzel" as term for roots of a number and "Lösung" for the roots/solutions of polynomial equations. It's much easier to mix up things if they are given the same name.
It is possible if -1 is replaced wIth i^2 and is carried over to m as i^4: 3 - sqrt( (2* (i^4*23/2) +2) = 3 - sqrt( (2* (i^4*23/2) +2i^4) = [added i^4 multiplier to 2 as i^4 is 1 and 2*1=2 3 - sqrt( (2* (23/2) +2) *i^4) = 3 - sqrt( (23+2) *i^4) = 3- sqrt( 25 * i^4 )= 3 - (5*i^2)= 3 - (-5)= 3+5=8 This solution is a bit long-winded and relies heavily on i, but it does work.
My hangup with this approach is i^4 = i^2 x i^2 = -1 * -1 = 1. So 3 - sqrt (25*i^4) can just as easily be simplified to 3 - sqrt(25), or 3 - 5, which does not equal 8.
First, you forget that the square roots of 25 are 5 and -5, and since 3 - -5 = 3+5 = 8, that does work. Second, at no specific time during the equation are you required by the rules of math to resolve i^x powers to 1 or -1, so requiring that it be resolved before you take into account square roots halfway through the equation without dealing with powers as well is a bad argument.
@@brianwilson14 Interesting argument. Focusing in on the fifth equation "3 - sqrt(25*i^4)". So let's say we wait until the square roots are first resolved. √(25*i^4) is the same as √25 * √i^4. We can then say this resolves to +/5 * i^2, or +/- 5 * -1. But, if I substitute the proposed answer m=8 into the original equation 3 - √(2m+2), my calculator shows that 3 - √(2*8+2) equals -1.23. The question is what is the value of m so that 3 - √(2m+2) = 8.
This feels like one of those things that comes up differently depending on the level of math that has been taught. The entry point is people only think about the principal square root, but as things like polynomial equations are introduced, you start to get "oh yeah... theres a secondary root". Then again, i was always getting into trouble on tests for having read ahead and applying theorems that hadn't been introduced in class yet.
If I had to grade this video, and most of the comments, on a gray-scale of right to wrong, I'd give it a dark gray. For the greenhouse-plant-algebra of high school, you can go ahead and only allow principal roots (and if you want to pass exams, you better beware), but that may not give you _working_ solutions in engineering and physics. In another reply here, I mentioned a 1899 Algebra (school) textbook that was a little more nuanced, where the telling phrase appears on page 241 "... _if other than principal square roots _*_be admitted_* ..."
@@-danRNah, man. You're just confusing two different notations: x^0.5 and √x . The former refers to both roots while the latter really just refers to the principal root. Otherwise, why is it that the quadratic formula has a ± in front of the √? It would be redundant if, as you claim, √ refers to both roots in "more advanced mathematics".
It’s ABSOLUTELY TRUE that the square root of 25 is plus or minus 5 HOWEVER The definition of the mathematical symbol √ is NOT simply the square root. The MATHEMATICAL DEFINITION of the symbol √ is “the principal square root”. ie. the positive square root. That’s why mathematical formulas like the solution to the quadratic equation include the symbols ±√ Those two symbols used together are referring to both the positive AND negative values of the square root.
@@nedruss7040 No, 12.5 * i^4 - 1 is not a valid way to express a complex number. In fact, 12.5 * i^4 - 1 = 11.5, which is a real number. All complex numbers can be expressed in the form *r (cos θ + i sin θ),* where r ≥ 0 and −π < θ ≤ π and where principal square root is *√(r (cos θ + i sin θ)) = √r (cos θ/2 + i sin θ/2)* Since −π < θ ≤ π, then −π/2 < θ/2 ≤ π/2, But cos θ/2 ≥ 0 for values in this interval, then principal square root of any complex number will always a non-negative real component. So for some complex number m, √(2m + 2) = a + bi, where a and b are real and a ≥ 0. Therefore √(2m + 2) = −5 cannot have any solution real or non-real complex.
@@nedruss7040 My front yard is 34 by -19 feet? Domain is always important in any real world math. I have never had any real world use for imaginary numbers, or otherwise out of the domain for the actual answer, except as a flag for "not this solution."
@@MarieAnne. The problem arises when there is one imaginary solution and one real one. That was the case in the only time I used advanced math skills outside a classroom. I was installing an Automatic Direction Finder (ADF) in a small plane and the specifications for the effective height (simple equation) of the wire antenna and capacitance of the antenna to the airframe (another simple equation) could not be met without a series capacitor at the receiver. The solution was the real root of a complex number; the imaginary root was useless.
@@flagmichael The fact that you've not encountered it doesn't mean that it doesn't exist. And BTW, "34 by -19 feet" would still be real, just negative.
I have a master's in math. I saw that was no solution immediately because the negative square root of (2m + 2) is not 5. When you put something inside a square root radical with no sign in front of the radical, it is understood "Take the positive square root of that something". I learned that in high school. When dealing with radical equations like the ones above, always test the answers!
You are limiting yourself to a real number solution. There is no real number solution, just as there are no integer solutions nor rational number solutions. However there are complex number (real plus imaginary) number solutions.
PEMDAS, but backwards because we are solving for a variable. You need to take care of Multiplication, Division, Addition and Subtraction BEFORE you deal with the square root (which is an exponent (1/2)). If you divide both sides by -1 you see that you need the square root of (2m+2) to equal a -5 (negative 5) we see we have a problem.
Technically this is HOW you evaluate something but abiding by this way… isn’t how it works… everything is solved in the same order but we must add stuff to find it.
Step 1 of resolving any equation with a variable in a radical or under a division bar is to remeber your domain and range. The original questions ask 3 minus [what] = 8. There is only 1 value in the [what] that works, -5. But the range on the sqrt function excludes negative values, so no further math needed. No sol.
This is a game of chess and checkers at the same time on the same board. Everyone adopts the rules they want. The square root is defined as non-negative. Not because those who adopted such a definition were stupid, but just to avoid unnecessary ambiguities. The consequences of leaving such ambiguity are much greater than the apparent deprivation of an "alternative solution". If a root can be both positive and negative, then every real number is equal to 0. Proof? Here you go: 2*sqrt(1)=sqrt(1)+sqrt(1), Now, first sqrt(1)= +1, second sqrt(1)= -1 - why not? 2*1 = +1 + (-1) 2 = 0 Now replace "1" with square of any real number - this way it will turn out to be equal to 0. Is this still mathematics?
Месяц назад+5
Thank you for explaining in 10 seconds what the video failed to explain in 17 long minutes!!
TI 89 yields "false". Multiplying both sides of -√(2m + 2) = 5 by -1 gives √(2m +2) = -5. A square root will not give a negative real solution. Negative 5 a solution to √25? If so, that would mean √(-5)√(-5). Bzzt! Wrong.
Your explanation of why you can’t use-5 by demonstrating a quadratic equation is weak for me I think you mean that if there is not a root (unknown) on either side, you cannot introduce 2 solutions (because that would be extraneous) according to wikipedia, by the way, a quadratic equation can have a single solution “double root” please clarify
A double root isn't a single solution, it's the same solution, repeated. Think of a parabola that crosses the x axis twice - two solutions. Now move it upwards until it just touches the axis - only one answer, because both roots are the same. Move it further up and there are no real solutions (two complex solutions).
the equation x^2 = 25 has two real solutions +5 and -5. (plug them in, they work) the equation x^2 -11x + 30 = 0 (x-5)*(x-6) = 0 has 2 solutions: 5 and 6 the equation x^2 -10x + 25 = 0 (x-5)*(x-5) = 0 has a "double root" of 5 the equation x^2 = (-25) has no real solutions -- there is no real number x, that when squared equals ( -25).. But, the complex number "i" is defined so that i^2 = -1. so (5*i)^2 = 5^2 * i^2 = 25 * (-1) = -25 ... so x = 5i is a solution. also (-5*i)^2 = (-5)^2 * i^2 = 25*(-1)= -25 ... so x = -5i is also a solution the equation "z = √(-25)" has no real solutions; the function √x is not defined for negative x. BUT √x IS defined for complex x and √(-25 + 0*i) = 5i (the principal square root) not ±5i. Note √25 = 5 (also the principal square root) not ±5 But wait ... aren't (-25 + 0*i) and (-25) equal ? Nope. (-25 + 0*i) is a complex number - think the ordered pair [-25,0] but (-25) is just a real number. that help ?
love your videos. I'm an engineer in my 50's and surprised how much basic math I have forgotten. Q: When m=0 why not further defined as empty set. Would like m = null or or other than a value of zero.
At first glance, there can be no (real!) solution to this equation, because a quick rearrangement of it yields: 3 - 8 = -5 = √(2m+2) which is impossible, because a radical must always be ≥ 0. Bottom line: *No Solution.* Fred
In practical situations this equation would arise while solving a problem with +/- in front of the radical and the steps you illustrate would eliminate the case with the - sign.
I used this rearrangement in my head to come up with a solution but forgot about the possibility of introduction of extraneous solutions. Oops! It's been over 40 years since I've been to Algebra class, perhaps I had a senior moment!
I agree with that method of conviction. However, explain to me how the way the square root equation is written determines whether there is a negative answer.
The easiest way to see that there is no real solution is to move everything to the rt and graph it. Y = sqrt(2x+2) + 5. (Have to change the m to an x for the calculator) Then graph. Notice that the graph never touches or crosses the x axis.
@@omarjette3859 "Well, for x=0 then f(x) = sqrt(2)-5 = -3.585" Where did you get the -5 from? The comment your replied to explicitly wrote +5- "so maybe infinitely close to x=0" ??? Where did you get that from?!?
If I was to apply this to an engineering problem such as “the difference of pressure of a pipe from a large diameter to a small diameter is 3-(2m+2)^0.5 =8 (Excuse the nomenclature, my phone doesn’t have a square root symbol) where m is the pressure difference in psi, then using (25)^0.5 =-5 is definitely valid as a negative pressure difference shows a drop across the pipe restriction. In fact, that there is just one answer of 11.5 is correct.
You don't even have to go to engineering just a pure mathematics perspective. The fact is that negative 5 is a square root of 25 it may not be the Principal square root but it is a square root it is a solution.
X^0.5 is not the same as √x . The √ symbol denotes the principal root while the exponentiation notation denotes all the roots. That's why there's a ± symbol in front of the √ in the quadratic formula. After all, if √ really did refer to all the roots, then the ± would be redundant, right?
@@misterroboto1 if √ does not include all roots why does the quadratic formula bother with ± Why not just do the quadratic equation like this: x = (-b + (b^2 - 4ac)^0.5)/2a It would be the same thing as x = (-b ± √(b^2 - 4ac))/2a The result is the same thing, including the fact that you come up with only two unique values. What the ± does is clarify the fact that there are two unique values for x.
You are totally wrong. This has nothing at all to do with being a real-world problem or not. If you wrote down an equation like 3-(2m+2)^0.5 =8 and claimed that this described a real-world problem with an actual solution, you are simply describing the real-world problem in a wrong way! This equation can _not_ describe a real-world problem with a solution, because this equation has no solution.
Please don't rely on the Magic Decoder Ring of PEMDAS/BODMAS. If you are working on somebody else's problem you never know whether it was written correctly (if the expression does not parenthesize enough to be clear it was sloppy from the outset. (Aced my math ACT test, scored second in the school in the MAA competition in my senior year of high school. I have tutored friends and family in math off and on since John F Kennedy was still the US president.
Thank you very much for your video. I made exactly that mistake and I noticed that it cannot be the correct solution. But please: Don't tell me again and again that I made a mistake for so many minutes of your video. Please tell me up front WHAT mistake I made.
Good video, thank you. A couple comments. When showing the algebra that leads to x=+/-5 you should include the step in between with the absolute value being the result of the square root of a squared variable. I take that opportunity to explain that we have to consider that if an unknown number was chosen, we cannot be certain what the sign was, and ABS(x) is how we admit that, while the sqrt(25) is 5, always and forever, due to PRT as you noted clearly. The other thing I would have mentioned is that you should avoid squaring negatives away by first moving it to the other side of the equation and observing that allegedly Sqrt(2m+2)=-5 which can't be true, because again the result of a sqrt is always positive.
YES. When showing someone how to solve by completing the square, I also show the "missing" step with absolute value. As you say √(a²) = |a| so we can get solution to a quadratic equation as follows: x² − 2x − 3 = 0 x² − 2x + 1 = 4 (x − 1)² = 4 √((x − 1)²) = √4 |x − 1| = 2 x − 1 = ±2 x = 1 ± 2 x = 3 or −1
Should start by checking the radicand for non-permissible values: ie 2m + 2 must be zero or greater or m must be greater than or equal to -1 which identifies -5 as an extraneous root. Cheers.
This is an argument of definitions more than how math actually works. Which is important when solving written equations because everyone has to be on the same page as to what the written expressions actually mean.
So, basically, the square root of (2m+2) must equal -5 for the original equation to work. This is what I was thinking… looks like my algebra is quite rusty. Thanks for the explanation. Makes total sense that there is no answer.
Forget the square root part, just looking at it and the problem is getting a bigger number than 3 while subtracting from 3. Nothing about it makes sense to my college algebra semester understanding.
I would expect the solution is m=(25/2)i - 1, because 2m+2 = 2x((25/2)i-1)+2 = (25i -2)+2 = 25i. With sprt(25i) = sprt (25) x sqrt(i) = 5 * (-1)= -5 the equation is fullfilled.
√(i) ≠ −1. You're confusing it with √(−1) = i In fact i = cos(π/2) + i sin(π/2), so √(i) = cos(π/4) + i sin(π/4) = 1/√2 + i/√2 There are indeed no complex solutions to this problem. Complex solutions occur when we have an equation of the form (ax+b)² = negative value but not when we have an equation of the form √(ax+b) = negative value ² ³ − √ ° ∠ △ × → ~ ≅ π ≈ αβθ ± ≤ ≥ ⁻¹ ⇒ ₁ ₂ ≠ ╪ ·
That’s something I can’t understand. If someone thinks that sqrt(16)=+/-4 and sqrt(9)=+/-3, what will be the result for sqrt(16)+sqrt(9)? Will it be either 7 or -7 or 1 or -1?
@@JacquesLafont There is a workable solution if the domain of the answer is complex numbers or the numerical solution is in the realm of imaginary numbers (sometimes that is true).
For me, what helps me understand what's going on is to hypothesize a simpler example. What is the value of x where √x = -1? Well, let's substitute the imaginary number i so that i² for -1, giving √x = i². Squaring both sides would give x = i^4, or x = i² * i², or x = -1 * -1 = 1. But substituting back into the original equation, √1 does not equal -1, it equals 1. So, there is no solution.
I didn't see how there could be a solution, but since you said to go ahead and use your calculator to get one, I came up with 23/2 and joined the majority.
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Squaring roots can introduce extranous solutions which is why you always have to check after solving an equation this way. This is exactly what he is showing here.
@@markmurto That gets into splitting hairs in the real world, particularly in AC electrical situations and in electronics. Half a century ago tunnel diodes were a darling of radio electronics; if biased properly they exhibited *negative* resistance over a small operating range and the speed was effectively at the speed of light. They would still be a major factor in some electronics today if they didn't have a problem of short life (months or less, depending on the service). In the domain of real numbers (most of the world around us) you are right, In some of the darker paths of exotic tech, you are left behind.
Another way to really emphasize the principle square root as a positive value is using functions. In order to pass the vertical line test, you have to have separate functions, f(x) = sqrt(x) and g(x) = -sqrt(x).
It's been a while since taking algebra and I think I vaguely remember this lesson. I think spending a lot of time with quadratic equations can cause one to forget this more basic lesson. When squaring a number, you have to consider both the positive and negative values that work. When taking the square root, unless otherwise denoted, such as in the quadratic equation with + - square root, it is only the positive root and in this case the positive root doesn't work as a solution and the equation doesn't work as a quadratic.
The root symbol is for the "principal square root". The the principa square root is a mathematical function which is defined as: "which positive number must be multplied with itself to get the positive number under the root symbol?". This function is only defined for the positive x/positive y-quadrant of the cartesian coordinate system. You as a human know that (2)² and (-2)² both result in 4. But the the principal square root can only reverse the first expression, not the second. That's why you often see ± in front of formulas containing square roots.
- SRQT(2m+2) = 5 is the same as: SRQT(2m+2) = - 5 The square root of any real number is a positive real number. Therefore the answer is an complex number.
Yes, I was one of those doing algebra in the '60's! Retired from Power Industry, but now have Parkinson's Disease and was looking for a channel like yours to try to keep sharp! I saw the null set right away! However, I do not recall having ever heard of the "Principal Square Root" before. Somehow, I learned the concept because -5 for the answer DID NOT occur to me. I remember that quadratic equations result in the answer being plus or minus, but I don't remember how to check those results. Unless you substitute in the original equation and discard the extraneous result.? Not sure. I will be sure to follow up with your other episodes! Semper Fi Bob
I think the main issue is that people seem to remember that when resolving a square root you introduce a "+-" but that's not the case. The "+-" does not come from solving the quare root on the right side but it comes from the left side. x² = 25 means x is a variable and when taking the quare root of x², this is where we introduce the two solutions, not when resolving the sqrt(25). So in some sense it would make more sense to write x² = 25 sqrt(x²) = sqrt(25) +-x = sqrt(25) +-x = 5 => x1 = +5, x2 = -5
Yes, I was wondering through RUclips. I found your videos. I’ll try to solve some of your equations most of them right but blew a couple others. I have to tell you that you are an excellent instructor. I struggled with math in high school, l failed algebra, I failed geometry. Sometimes I failed them twice, for my last year and a half of high school I transferred to a different school. I needed to pass geometry which I had failed twice. Praise the Lord for my new instructor it was a whole new game, passed geometry one and geometry two with A’s. I don’t believe my former instructors were bad, but I think Rob knew how to teach me. I tell my friends, who have children who struggle with math I’ to change to a different teacher if possible or find a tutor who clicks with their child. I went to college and luckily had great instructors and TA’s until integral calculus, he was not a good fit. Now days i use my algebra and geometry to make quilts. I plan on watching your videos to add math to my crosswords and puzzles to keep my brain moving. I am 75 years old. Thank you so much.
That is when you hope to step on the green carpet on the sidewalk and find it is just a floating mat of algae. You say "bah!" to complete the meme. Say it!
This is great but I'm confused why I couldn't take this path: ... 3 - sqr(25) = 8 ==> -sqr(25) = 5 ==> sqr(25) = -5 ==>(sqr(25))^2 = (-5)^2 ==> 25 = 25
Tidying up, I get -5 = sqrt(2m+2). Since the radical sign implies a positive square root, there is no solution (in real numbers) because the left side is negative. [ Final answer! ] When I check for imaginary roots, I square both sides, and get (-5)^2 = absval(2m+2). Split this up into two possibilities, 25 = 2m+2 and 25 = -(2m+2). This ends up with m1 = 23/2 and m2 = -27/2. Neither of these is equal to 8, so there is no complex solution either. P.S. I also taught high school algebra for 30 years, but I retired 20 years ago, so I'm a little rusty!
When you evaluate a term, you have to resolve a radical before applying a sign to it; there are implicit parentheses around the radical expression. In this example, you can subtract 8 from both sides, giving -5 - (radical expression) = 0 upon which you can add the radical expression to both sides, giving -5 = (radical expression), and you know immediately that won't fly for real numbers. This is somewhat similar to the kinds of multiple-root fallacies you can get in calculating an internal rate of return (IRR) for a business project. It's a pitfall people too frequently fall into in trying to get numbers to sell a project to management, and, too often, neither those who present the numbers nor the management receiving them understand how such unrealistic predictions occurred until the unrealism comes jome to roost. Moral: don't use IRR to sell something; instead, calculate the range of credible NPVs.
A lot of emphasis there on correct maths, which is fine .. but wouldn't it be easier simply to take the original equation, remind the student that the square root symbol means (by definition, e.g. the requirement to append a +- in the QE formula), the POSITIVE root and so the equation is impossible.
Min 13:38 has little to do with sqrt itself. Rather the fact that x^2=25 x^2-25 = 0 (x+5)(x-5)=0; thus the solution is x=+5 or x=-5. You simply use the fact that sqrt(25)=5 > 0 in the factorization, but the double root comes from the fact that there are 2 factors (x+5) and (x-5).
Another way of looking at it is, if you're going to say SQRT25 = ±5, then you can't just choose -5 arbitrarily to make the game come out. Logically, *both* +5 and -5 must satisfy the original equation and this is clearly not the case.
Well, that didn’t go where I expected (but I’ve always considered myself fairly mediocre at algebra). Before playing the video, I just evaluated the thumbnail in my head, started moving things around, and quickly came to the expectation that “the thing people forget“ is that “you’re going to need i” - so I was surprised the answer wasn’t “5i”
I now know why I was so bored doing Mathematics at school, teachers who like the sound of their voice better than giving a concise reasoning in as few words as possible.
Well, the equation has no solution if you stick to a rigid definition of a square root as something always positive, but if you allow yourself to let a negative number also be a square root, it indeed has a solution. What it also shows is that mathematics is not entirely without faults.
No. It just shows you didn't understand square roots. Apart from that, in mathematics one should always stick to the definition (call it "rigid" if you want) and not allow yourself to let other things be something which they are not.
@@WK-5775 I understand that thing very well. But this implies that I also understand that what is wrong in the first place, is math itself, as it is formulated.
Thank you. Great review. I’ll check out your general review of math skills. Took college level algebra and calculus, but never had the same comfort with calculus that I did with algebra, and was probably why I didn’t enjoy physics. Maybe I can move beyond that now that I’ve retired and want to better understand physics beyond mechanics and electricity.
A key thing to notice is the negative in front of the square root after moving the 3. Having that then equal a positive number on the other side of the equation should be an alert to you.
Step two is either sqrt(2m+2) = -5 or you can use your step two and make this step three. Once you get to this point, you have a square root equal to a negative number, but square root always means ONLY the positive square root. There is no square root that can equal -5; therefore, there is no solution.
Isn't (-5)*(-5)=25? And that means the square root for 25 is +/-5. I was always tought by all teachers that there can be two square roots, a positive and a negative, if not stated explicitly before the square root sign with a plus-sign. I have never ever heard of "principal square root" and I don't get why it is used here. Still see it is two possible answers with either 23/2 OR no solution. Maybe this is a new convention that wasn't taught in the 80:s? (or maybe no in Sweden?). I mean many others writing in here who were taught it decades ago seems to think the same thing.
There is an answer It is 11.5 x i where i is defined as the square root of -1. When I took Algebra, we were taught imaginary numbers when we learned how to solve quadratic equations.
I can remember from maths lectures during my uni studies of chemistry we should ALWAYS keep in mind that squaring is not an equivalent transformation. Bad things can happen...! 😉
3 minus -5 = 8 Given that a square root cannot produce a negative result, the solution is null. Or you could just graph both sides; the graph stops as m approaches -3 and never intersects.
Quick question in terms of symbology: does (25)^1/2 mean the same as square root of 25? I don't don't know how to type a square root symbol on my phone but that's what I meant. In other words does using the 1/2 exponent mean both +5 and -5 are solutions because you are not using the square root symbol?
(25)^1/2 is the same thing as √25. So too is 25^1/2 or 25^0.5 or 25^(1/2) or 25(^1/2) or (25)(^1/2), but you'll rarely see two pairs of parentheses. Some viewers use sqrt 25. Some use V25 but that's not very nice. Whether one form or another is used, it's still square root.
All you need to do is to multiply each side of the original equation by (-1) in order to get rid of the negative sign in front of the radical. You will get an identical equation but will not have to deal with the confusing negative sign. Thus, you will get [-3 + radical = - 8], which leads to [radical = - 5], which is impossible for real numbers (unlike for imaginary ones).
If you are stuck with the idea that the solution set has to include both 5 and -5 for the square root of 25, keep in mind that the reason quadratic equations result in a solution set of two possible values for the variable is because _both_ values will result in a correct solution. In this case, one of the two possible values does _not_ result in a valid solution. This means that there is no valid solution set. And yeah, I was stuck m = 11.5 too.
You've essentially demonstrated that the solution the majority of people will reach is correct. I challenge you to demonstrate how this has been a problem in industry.
Just a sidenote: At 12:37, the resolution to the equation shown at the right of the screen yields the correct result but there's a problem on the second step. We know that √x²=|x|. Yet you wrote √x²=±√25 which is wrong. It should be: √x²=√25 Then |x|=5 And finally x=±5. By saying that √x²=±√25 you would be saying that √x² can be a negative number, which it cannot be. Just a small inaccuracy that somewhat contradicts the main point of the video. Now, that resolution to the equation may very well be there to demonstrate how some people may get that step wrong while solving for x, but as it is not stated anywhere on the video i suppose it's just a small mistake that made it's way into the video.
You made the mistake you are trying to avoid at 13:24. "So the square root of x squared is x" - that is not true. It is important to not reinforce this fallacy. sqrt(x^2)=|x| not x. |x|=5 => x=5, x>0 and -x=5, x
Once you get that the square root is a negative then you could only get a solution if you use complex numbers i.e. i in other words 2m + 2 = sqrt(5) * i Without using complex numbers there is no value that you can square to get a negative number.
I simplified question to 3-x=8 Seeing x had to be a negative and intuitively knowing you can't get a negative from square root knew something was wrong. First thought it might be imaginary but that didn't work out either so knew there was no solution. Did I just get lucky or was logic correct..
My algebra is a little rusty after 40 years, so I tried to look at this problem logically - not that algebra isn’t logical…..lol😂 Fundamentally, 3 minus anything cannot = 8 unless you’re subtracting minus 5 but Sq.Rt of (2m + 2) cannot be minus 5. It’s basically the same logic.
An introduction to *_Social Justice_* mathematics:
a = b initial supposition a*a = a*b multiply both sides by a a*a - b*b = a*b - b*b subtract b*b from both sides a*a - a*b + a*b - b*b = a*b - b*b substract and add a*b simultaneously on the left side a*(a-b) + b*(a-b) = b*(a-b) factor out (a-b) thrice (a+b)*(a-b) = b*(a-b) combine a+b a+b = b divide out (a-b) b+b = b use the first equation a=b 2*b = b reduce b+b to 2*b 2 = 1 divide out b 3 = 2 add 1 to both sides 4 = 3 add 1 to both sides again 5 = 4 ... and continue adding 1 to both sides 1 = 0 subtract 1 from both sides of 2 = 1 0 = -1 subtract 1 from both sides -1 = -2 subtract 1 from both sides again -2 = -3 -3 = -4 -4 = -5 ...
... -5 = -4 = -3 = -2 = -1 = 0 = 1 = 2 = 3 = 4 = 5... and so I have proved that all integers are equal!!
"undefined" and "no solution" are the correct ways to express it. What he wrote ("m = ∅") is incorrect. If you wanted to use set notation, you'd have to write "{m} = ∅".
Mathematics needs a symbol for the operation that returns a positive and a negative value for the square root, in other words for the reverse of the square function. There needs to be an operation where the value for -5 (or some other number) squared can then be reversed to return the original value.
The result of the square root of 25 is the value of 5. If you square a positive 5 or a negative 5 you still get the same result of positive 25. That is why the square root of a number technically is +/- the magnitude of the result of the square root operation.
@@GaryBricaultLive Yes, I got confused. The square root of a number can be negative. I was thinking of the square root of a negative number, which is a different thing.
@@wernerviehhauser94 There is sometimes value in the negative aspect of a square root. If we are going to fall short, it may be helpful to know whether we can come up winners next time. It won't work this time, but may define a fix next time.
2m=1 I think. 9^1/2 - (2m+2)^1/2 =8 -2m+9-2=8 | -2m=8-7 | -2m=1 2m=1 m=-1/2 Now let's look at what you did lol na I screwed up combining the roots impossibly
I have an intuition that the +5 solution would draw a different shape geometrically than the -5 solution. I have no intuition that sqrt(25) does not include -5 as an answer.
Hey this is interesting. Without having watched the videos, here's some thoughts. √(2m + 2) = -5 can't be solved on the real plane, so it must be a complex something. It looks like the modulus of a complex number [Z_modulus = √(a^2 + b^2)] Z_modulus = √(2m + 2) = -5 Write that as the conjugate [Z_conjugate = a - ib] and you get the following I guess? Z_conjugate = √(2m) - i√2 = ??? Don't really know how to continue, or if these even remotely made sense or is useful.
When I invented Algebra back in ‘68 I didn’t realize how confusing it could get. If I could go back and do it again I would leave out all the letters and just stick with numbers.
Wow, really wonderful observation there. You mean to tell me that I didn’t actually invent Algebra? You mean it was actually developed over 3000 years ago? Thank goodness there are sharp people like you around to make sure jokes aren’t taken literally. I’m sure all of the people who thought a random RUclips commentator invented algebra are grateful for your wisdom.
Thanks; What if the positive sqrt gives the correct answer, but the negative sqrt does not (i.e. if 3-sqrt(2m+2) = -2 ) Do I also conclude there is no solution?
That the sqr is only positive is only an arbitrary definition, not something fundamental. In the real world you should take it as a possibility that the writer intended +- sqr. Especially if they state that there is a solution.
sorry to break this to you, but this has nothing to do with "real world applications" and all with "didn't really understand how numbers work". If the author of the question intended a +- square root, he was in error.
Yes and no. The only time I used my advanced math skills in the real world was to calculate the value of a series capacitor I had to put in line with a wire antenna to bring the capacitance down to spec. If the value had been imaginary I could have looked at shortening the wire antenna because it would not work this time.
An important general point is this: when you square both sides of an equation, your new equation can have more solutions
than the one you started with. Ex: x = 2 has one solution, x^2 = 4 has two solutions. When you solve an equation after having
squared both sides of the original, your new equation may have one or more "false roots". Thus, it is necessary to take your
bottom line "possible solutions" back to the original to see which ones don't work.
Yes, thank you. A teacher pronouncing that "only the principal square root matters in this case" tells the student nothing useful whatsoever. I already knew the answer, and this video almost sucked that knowledge out of my brain.
That's true, but it's not really relevant here.
Going through a whole process of solving the equation (which involves squaring both sides) only to find that the solution doesn't work when you plug it back in to the original equation is a complete waste of time (not to mention a failure of understanding) when we can see by inspection that the equation is impossible.
This is true whenever you apply a non-injective function to the beginning equality.
For example,
x=2
If we apply f(x)=x+1/x to both sides, we get
x+1/x= 5/2, which has 2 solutions. Mainly,
x_1= 2 and
x_2 = ½
Wow
Thanks for that. I was very frustrated after sitting through the lesson, because he didn't explain clearly (at least to me) what had gone wrong. I thought, if you do the same thing to both sides of the equation, the resulting equation is guaranteed to be correct. If you follow all the rules and get the wrong answer.... wha??? That was the feeling I was left with. But your general point cleared that up for me. It was just what I needed. The teacher made it sound like the problem was the definition of a symbol, which seems like just a matter of convention, an arbitrary rule, not a mathematical truth. That didn't explain how you can get the wrong answer by following a rule that should guarantee the preservation of truth. Your example showed that the problem of extra roots didn't depend on the radical symbol. It was a consequence of the squaring operation.
This is a really important thing to understand. I don't know the exact percent of students who don't get it, but from decades of tutoring, I know it's high.
From the comments there are some adults who are also confused by it. (Which is okay.)
If you ask the solution of x² = 4, it's x ∈ {-2, 2}. But if you ask what is the solution of √4 = x, then the solution is ONLY x=2. The square root symbol means ONLY the positive square root, or what they call the "principal square root."
Start with:
x² = 4
Take square root of both sides:
±√x² = ±√4
Simplifying:
x = ±2, or x ∈ {-2, 2}
This is why in the quadratic equation, they use the ± symbol - because the square root symbol by itself refers to ONLY the positive (principal) square root and since the equation needs to account for BOTH square roots, they have to include the plus or minus sign.
Nostalgia here!
That's a very clear explanation. Your comment deserves to be pinned, as it would help a lot of people.
The problem here is that there is a distinction between mathematics meant to have single answers and mathematics used to solve actual real world problems.
Except that according to Ozford website, sqrt of X is plus AND minus UNLESS a positive or negative sign is applied to the sqrt symbol.
Except for 'checking the answer with the original problem', you basically explained the whole 17+ minutes video in less than a minute of reading. Thank you, well done.
The first step you have to take is to spell ALGEBRA correctly.
Math skills/language skills....and never the twain shall meet.
lol... spotted that straight away
you so wierd
The second step is to spell WEIRD correctly.
@@luisgq2358 Weird
In this case it would be informative to move the - to the numerical side of the equation. It would yield a sq root equalling a negative number which is an obvious thing to look for.
That was basically going to be my comment. You can tell right from the beginning that way that it has no solution.
Yes. Moving the sign immediately gives the solution (25/2)i - 1
Yeah, it never even occurred to me to square away the negative.
that’s how i looked at the problem from the get go
@@ozviking8052 That's not a solution. That gives 2m+2 = 25i.
But √(25i) does not equal -5.
I think you have the relationship backwards. It looks like you're thinking √i = -1, but that's not correct. The relationship is √(-1) = i.
3-√(2m+2)=8
√(2m+2)=-5
√x is the primary functional branch of the inverse relation of the function f(x)=x², with branch cuts such that f₁⁻¹(x)≥0 and f₂⁻¹(x)
The radical has a negative of an imaginary operator on it, Heaviside's j=±1 rotating signs with each operation that squares to -1. When we divide both sides by that, we rotate it into another operator that does not flip its signs and squares to +1. It eliminates the extraneous solution.
3-±√2m+2=8 → -j√2m+2=5 → √2m+2=5/-j rationalizes to 5j'=√2m+2 → 25=2m+2
Wikipedia states definition includes both positive and negative solutions…..
@@peppermann wikipedia also states that the radical denotes the principal square root, also called "THE square root" with a definite article, which is positive. The radical symbol is representing the square root FUNCTION, which takes the real numbers exclusively to the non-negative complex axes. A function maps every element of its domain to exactly one element in its codomain.
In general "a root" is a solution to a polynomial equation in one variable. An "n'th root of k" is a solution to the polynomial xⁿ-k=0. The roots of x²-4=0 are {2,-2}, but √4 is only 2.
You need to use imaginary numbers.
@@ProactiveYellow using a definite article to imply something significant. At first glance that seems to be practice that should be disavowed, but thinking back, a lot of statements in math also make quite heavy usages of grammar rules. But "THE" here really is a thing that's easy to miss.
I got "no solution" a slightly different way. I don't consider "-(sqrt(2m+2)) to be fully isolated, since there's a -1 being multiplied by (sqrt(2m+2)) there. So I divided through by the -1 to get sqrt(2m+2)=-5. At which point you already know there is no solution, since a sqrt can't give a negative answer.
Actually a sqrt can give a negative number. Without complex numbers you can't take the sqrt of negative number. - 5 * - 5 = 25 and 5 * 5 = 25 so the sqrt of 25 is +/- 5.
@@alexaneals8194 The problem here is you both are talking about different sqrt functions. You need to agree first which sqrt to use. One of them can give values in the form of negative real numbers and the other one is not allowed to do that.
@@evgtro8727 It's actually not which sqrt to use. It's the fact that they have introduced the concept of principal sqrt after I attended school. I took Algebra I in the 8th grade in 1978-1979 and Algebra II the following year 1979-1980. At that time there was no principal sqrt. The sqrt equation was assumed to include both positive and negative values. So, the roots to 25 would be +5 and -5. Even the math and engineering handbook that I have from before 2000 does not mention the principal sqrt. So, the concept is new to me. It makes sense since most people use calculators in math class and the calculator like the programming languages treat the square root as a function and not as an equation (which can return more that one value for a given input).
@@alexaneals8194 Ironically I was the 8th grade student exactly at the same years, 1978-1979 🙂
I was a math major in a university in 1970. I have NEVER heard of " principal square root." Mathmatics is centuries old, this is a new invented term.
What promotes the era is when teachers self contradict when they write on the board something like rad(4) = +/-2. That is never acceptable. In this series of math lessons I’ve seen that done frequently. You cannot ever justify writing that a radical is equal to plus or minus anything. There is no justification for claiming that. Doing that is what creates the problems with radicals.
You are wrong. Two negatives cancel so it is acceptable. -(2)^2 =4. The problem is the new math and it's insane bias against complex numbers at the prealgebra stage when it is required for real world applications using complex analysis.
We see more and more this kind of nonsense.
@@josephmalone253wtf? Who tought you math?
First: -(2)^2 = -4 . Put the "-" within the parenteses if you want it to be squared, too......
Second: the radical sign is DEFINED to be the principal root. Always has been. Therefore √4 = +2 and nothing else.
If you want to solve x^2 = 4, you HAVE to write x = +- √4 what can be simplified to +-2. Everything else is incorrect notation.
And if you knew ANY math at all, you would KNOW that even complex numbers do not solve this. The radical sign is defined the exact same way for complex numbers. Don't need to trust me - try Wolfram Alpha. It'll tell you that there is no solution.
@@wernerviehhauser94I think the problem relies on people not realizing the context of the operation. Yes the radical is defined as the principal root, which is positive but then people get confused in operations like solving for x^2 because then you have 2 roots. If it's just a number with no other context but no relation to anything else then the principal root is the answer.
But in this scenario I think the right approach was not to give in the urge to solve for m and just analyze the operations.
3 - (any positive number) =/= 8
@@marilynman I also assume that the culprit here is, apart from "New Math" and the educational system, the use of "root" for two distinct things: roots of a number and solutions of polynomial equations. Here in Germany, you don't find many people making this mistake since we have "Wurzel" as term for roots of a number and "Lösung" for the roots/solutions of polynomial equations. It's much easier to mix up things if they are given the same name.
It is possible if -1 is replaced wIth i^2 and is carried over to m as i^4:
3 - sqrt( (2* (i^4*23/2) +2) =
3 - sqrt( (2* (i^4*23/2) +2i^4) = [added i^4 multiplier to 2 as i^4 is 1 and 2*1=2
3 - sqrt( (2* (23/2) +2) *i^4) =
3 - sqrt( (23+2) *i^4) =
3- sqrt( 25 * i^4 )=
3 - (5*i^2)=
3 - (-5)=
3+5=8
This solution is a bit long-winded and relies heavily on i, but it does work.
Smart! But… you did the thing sqrt(-1) * sqrt(-1)
= sqrt(-1 * -1)
= sqrt(1)
= 1
Which you cannot do as that.
My hangup with this approach is i^4 = i^2 x i^2 = -1 * -1 = 1. So 3 - sqrt (25*i^4) can just as easily be simplified to 3 - sqrt(25), or 3 - 5, which does not equal 8.
@@penguincute3564 sqrt(-1)*sqrt(-1) = sqrt(i^2)*sqrt(i^2) = i*i = i^2 = -1
First, you forget that the square roots of 25 are 5 and -5, and since 3 - -5 = 3+5 = 8, that does work. Second, at no specific time during the equation are you required by the rules of math to resolve i^x powers to 1 or -1, so requiring that it be resolved before you take into account square roots halfway through the equation without dealing with powers as well is a bad argument.
@@brianwilson14 Interesting argument. Focusing in on the fifth equation "3 - sqrt(25*i^4)". So let's say we wait until the square roots are first resolved. √(25*i^4) is the same as √25 * √i^4. We can then say this resolves to +/5 * i^2, or +/- 5 * -1. But, if I substitute the proposed answer m=8 into the original equation 3 - √(2m+2), my calculator shows that 3 - √(2*8+2) equals -1.23. The question is what is the value of m so that 3 - √(2m+2) = 8.
This feels like one of those things that comes up differently depending on the level of math that has been taught. The entry point is people only think about the principal square root, but as things like polynomial equations are introduced, you start to get "oh yeah... theres a secondary root". Then again, i was always getting into trouble on tests for having read ahead and applying theorems that hadn't been introduced in class yet.
If I had to grade this video, and most of the comments, on a gray-scale of right to wrong, I'd give it a dark gray.
For the greenhouse-plant-algebra of high school, you can go ahead and only allow principal roots (and if you want to pass exams, you better beware), but that may not give you _working_ solutions in engineering and physics. In another reply here, I mentioned a 1899 Algebra (school) textbook that was a little more nuanced, where the telling phrase appears on page 241 "... _if other than principal square roots _*_be admitted_* ..."
Exactly what I was thinking. This all comes down to what level of math your studying and what crazy rules your teacher wants to add in.
@@-danRNah, man. You're just confusing two different notations: x^0.5 and √x . The former refers to both roots while the latter really just refers to the principal root. Otherwise, why is it that the quadratic formula has a ± in front of the √? It would be redundant if, as you claim, √ refers to both roots in "more advanced mathematics".
I'm sorry you went through that.
It’s ABSOLUTELY TRUE that the square root of 25 is plus or minus 5
HOWEVER
The definition of the mathematical symbol √ is NOT simply the square root. The MATHEMATICAL DEFINITION of the symbol √ is “the principal square root”. ie. the positive square root.
That’s why mathematical formulas like the solution to the quadratic equation include the symbols ±√
Those two symbols used together are referring to both the positive AND negative values of the square root.
I should probably write /sqrt in abs( ) bars from now on just to avoid this pitfall bc I can VERY easily see myself making this mistake otherwise
Oops, didn't see this comment, and I mentioned the same.
The square root indicates a positive real number.Therefore,this equation has no real solutions.
It does have a complex solution, though. m=12.5 * i^4 - 1.
2m =25*i^4 - 2
2m + 2 = 25 * i^4 - 2 + 2, or simplified to 25 * i^4
The square root of 25 * i^4 = 5i²
i² = -1
3 - ((-1) * 5) = 8
@@nedruss7040 No, 12.5 * i^4 - 1 is not a valid way to express a complex number.
In fact, 12.5 * i^4 - 1 = 11.5, which is a real number.
All complex numbers can be expressed in the form *r (cos θ + i sin θ),* where r ≥ 0 and −π < θ ≤ π
and where principal square root is *√(r (cos θ + i sin θ)) = √r (cos θ/2 + i sin θ/2)*
Since −π < θ ≤ π, then −π/2 < θ/2 ≤ π/2, But cos θ/2 ≥ 0 for values in this interval, then principal square root of any complex number will always a non-negative real component. So for some complex number m,
√(2m + 2) = a + bi, where a and b are real and a ≥ 0. Therefore
√(2m + 2) = −5 cannot have any solution real or non-real complex.
@@nedruss7040 My front yard is 34 by -19 feet?
Domain is always important in any real world math. I have never had any real world use for imaginary numbers, or otherwise out of the domain for the actual answer, except as a flag for "not this solution."
@@MarieAnne. The problem arises when there is one imaginary solution and one real one. That was the case in the only time I used advanced math skills outside a classroom. I was installing an Automatic Direction Finder (ADF) in a small plane and the specifications for the effective height (simple equation) of the wire antenna and capacitance of the antenna to the airframe (another simple equation) could not be met without a series capacitor at the receiver. The solution was the real root of a complex number; the imaginary root was useless.
@@flagmichael The fact that you've not encountered it doesn't mean that it doesn't exist. And BTW, "34 by -19 feet" would still be real, just negative.
I have a master's in math. I saw that was no solution immediately because the negative square root of (2m + 2) is not 5. When you put something inside a square root radical with no sign in front of the radical, it is understood "Take the positive square root of that something". I learned that in high school. When dealing with radical equations like the ones above, always test the answers!
I haven’t done math since high school, and even I saw the same thing!
I have one semester of college algebra. How do you subtract from 3 and get a bigger number. It doesn't make sense to me.
You are limiting yourself to a real number solution. There is no real number solution, just as there are no integer solutions nor rational number solutions. However there are complex number (real plus imaginary) number solutions.
@@gregsmith4102 no, there aren't. Please provide an example of a complex solution
. So what is the complex solution?
PEMDAS, but backwards because we are solving for a variable. You need to take care of Multiplication, Division, Addition and Subtraction BEFORE you deal with the square root (which is an exponent (1/2)). If you divide both sides by -1 you see that you need the square root of (2m+2) to equal a -5 (negative 5) we see we have a problem.
Not the Magic Decoder Ring of PEMDAS/BOMDAS again! That should be trashed in favor of using nested parentheses because it is unreliable.
Technically this is HOW you evaluate something but abiding by this way… isn’t how it works… everything is solved in the same order but we must add stuff to find it.
Step 1 of resolving any equation with a variable in a radical or under a division bar is to remeber your domain and range. The original questions ask 3 minus [what] = 8. There is only 1 value in the [what] that works, -5. But the range on the sqrt function excludes negative values, so no further math needed. No sol.
I like it!
This is a game of chess and checkers at the same time on the same board. Everyone adopts the rules they want. The square root is defined as non-negative. Not because those who adopted such a definition were stupid, but just to avoid unnecessary ambiguities. The consequences of leaving such ambiguity are much greater than the apparent deprivation of an "alternative solution". If a root can be both positive and negative, then every real number is equal to 0. Proof? Here you go: 2*sqrt(1)=sqrt(1)+sqrt(1),
Now, first sqrt(1)= +1, second sqrt(1)= -1 - why not?
2*1 = +1 + (-1)
2 = 0
Now replace "1" with square of any real number - this way it will turn out to be equal to 0.
Is this still mathematics?
Thank you for explaining in 10 seconds what the video failed to explain in 17 long minutes!!
Can't this be "solved" by making all +/- agree?
TI 89 yields "false". Multiplying both sides of -√(2m + 2) = 5 by -1 gives √(2m +2) = -5. A square root will not give a negative real solution. Negative 5 a solution to √25? If so, that would mean √(-5)√(-5). Bzzt! Wrong.
If he says everybody this must include himself by definition.
but √(-5)√(-5) = i√(5)i√(5) = i^2√(25) = -1 *5 = -5
Your explanation of why you can’t use-5 by demonstrating a quadratic equation is weak for me
I think you mean that if there is not a root (unknown) on either side, you cannot introduce 2 solutions (because that would be extraneous)
according to wikipedia, by the way, a quadratic equation can have a single solution “double root”
please clarify
A double root isn't a single solution, it's the same solution, repeated. Think of a parabola that crosses the x axis twice - two solutions. Now move it upwards until it just touches the axis - only one answer, because both roots are the same. Move it further up and there are no real solutions (two complex solutions).
the equation x^2 = 25 has two real solutions +5 and -5. (plug them in, they work)
the equation x^2 -11x + 30 = 0 (x-5)*(x-6) = 0 has 2 solutions: 5 and 6
the equation x^2 -10x + 25 = 0 (x-5)*(x-5) = 0 has a "double root" of 5
the equation x^2 = (-25) has no real solutions -- there is no real number x, that when squared equals ( -25).. But, the complex number "i" is defined so that i^2 = -1. so (5*i)^2 = 5^2 * i^2 = 25 * (-1) = -25 ... so x = 5i is a solution. also (-5*i)^2 = (-5)^2 * i^2 = 25*(-1)= -25 ... so x = -5i is also a solution
the equation "z = √(-25)" has no real solutions; the function √x is not defined for negative x. BUT √x IS defined for complex x and √(-25 + 0*i) = 5i (the principal square root) not ±5i. Note √25 = 5 (also the principal square root) not ±5
But wait ... aren't (-25 + 0*i) and (-25) equal ? Nope. (-25 + 0*i) is a complex number - think the ordered pair [-25,0] but (-25) is just a real number.
that help ?
love your videos. I'm an engineer in my 50's and surprised how much basic math I have forgotten.
Q: When m=0 why not further defined as empty set. Would like m = null or or other than a value of zero.
At first glance, there can be no (real!) solution to this equation, because a quick rearrangement of it yields:
3 - 8 = -5 = √(2m+2)
which is impossible, because a radical must always be ≥ 0.
Bottom line: *No Solution.*
Fred
In practical situations this equation would arise while solving a problem with +/- in front of the radical and the steps you illustrate would eliminate the case with the - sign.
Rearrange the terms and you get √(2m + 2) = -5. The square root is always positive, so this has no solutions.
I used this rearrangement in my head to come up with a solution but forgot about the possibility of introduction of extraneous solutions. Oops! It's been over 40 years since I've been to Algebra class, perhaps I had a senior moment!
I agree with that method of conviction. However, explain to me how the way the square root equation is written determines whether there is a negative answer.
But they are trying to subtract from 3 and trying to get a bigger number. Am I missing something?
This is equivalent to Gödel respecting “This statement is false” .Both are lies, the work of The Deceiver
@@benprice3586 sqrt 2m = -7 (i.e. -5-2), or we can wrote it as sqrt m = -7/2. No real root.
The easiest way to see that there is no real solution is to move everything to the rt and graph it.
Y = sqrt(2x+2) + 5. (Have to change the m to an x for the calculator) Then graph. Notice that the graph never touches or crosses the x axis.
Well, for x=0 then f(x) = sqrt(2)-5 = -3.585... so maybe infinitely close to x=0. Could we say "at the limit"?
@@omarjette3859 "Well, for x=0 then f(x) = sqrt(2)-5 = -3.585"
Where did you get the -5 from? The comment your replied to explicitly wrote +5-
"so maybe infinitely close to x=0"
??? Where did you get that from?!?
If I was to apply this to an engineering problem such as “the difference of pressure of a pipe from a large diameter to a small diameter is 3-(2m+2)^0.5 =8
(Excuse the nomenclature, my phone doesn’t have a square root symbol) where m is the pressure difference in psi, then using (25)^0.5 =-5 is definitely valid as a negative pressure difference shows a drop across the pipe restriction. In fact, that there is just one answer of 11.5 is correct.
Absolutely agree. It is foolish to ignore the negative value of square roots. To say 11.5 is not a solution is just plain wrong.
You don't even have to go to engineering just a pure mathematics perspective. The fact is that negative 5 is a square root of 25 it may not be the Principal square root but it is a square root it is a solution.
X^0.5 is not the same as √x . The √ symbol denotes the principal root while the exponentiation notation denotes all the roots. That's why there's a ± symbol in front of the √ in the quadratic formula. After all, if √ really did refer to all the roots, then the ± would be redundant, right?
@@misterroboto1 if √ does not include all roots why does the quadratic formula bother with ± Why not just do the quadratic equation like this:
x = (-b + (b^2 - 4ac)^0.5)/2a
It would be the same thing as
x = (-b ± √(b^2 - 4ac))/2a
The result is the same thing, including the fact that you come up with only two unique values.
What the ± does is clarify the fact that there are two unique values for x.
You are totally wrong. This has nothing at all to do with being a real-world problem or not. If you wrote down an equation like 3-(2m+2)^0.5 =8 and claimed that this described a real-world problem with an actual solution, you are simply describing the real-world problem in a wrong way! This equation can _not_ describe a real-world problem with a solution, because this equation has no solution.
i'm an adult that's brushing up on my math skills. i find your classes interesting and helpful. thanks.
Please don't rely on the Magic Decoder Ring of PEMDAS/BODMAS. If you are working on somebody else's problem you never know whether it was written correctly (if the expression does not parenthesize enough to be clear it was sloppy from the outset. (Aced my math ACT test, scored second in the school in the MAA competition in my senior year of high school. I have tutored friends and family in math off and on since John F Kennedy was still the US president.
I write the square root of x^2 as |x|=5
Valid, but requires splitting the path in two from there on - one for 5 and one for -5. Remember the outcome is one OR the other.
@@flagmichael That is what the absolute value implies no? Guess what. I don't leave it as |x|.
Thank you very much for your video. I made exactly that mistake and I noticed that it cannot be the correct solution.
But please: Don't tell me again and again that I made a mistake for so many minutes of your video. Please tell me up front WHAT mistake I made.
It's starting to look like John takes pleasure in seeing his students make mistakes.
Good video, thank you. A couple comments. When showing the algebra that leads to x=+/-5 you should include the step in between with the absolute value being the result of the square root of a squared variable. I take that opportunity to explain that we have to consider that if an unknown number was chosen, we cannot be certain what the sign was, and ABS(x) is how we admit that, while the sqrt(25) is 5, always and forever, due to PRT as you noted clearly. The other thing I would have mentioned is that you should avoid squaring negatives away by first moving it to the other side of the equation and observing that allegedly Sqrt(2m+2)=-5 which can't be true, because again the result of a sqrt is always positive.
YES. When showing someone how to solve by completing the square, I also show the "missing" step with absolute value. As you say √(a²) = |a| so we can get solution to a quadratic equation as follows:
x² − 2x − 3 = 0
x² − 2x + 1 = 4
(x − 1)² = 4
√((x − 1)²) = √4
|x − 1| = 2
x − 1 = ±2
x = 1 ± 2
x = 3 or −1
Should start by checking the radicand for non-permissible values: ie 2m + 2 must be zero or greater or m must be greater than or equal to -1 which identifies -5 as an extraneous root. Cheers.
Got the error directly: the r is missing!
This is an argument of definitions more than how math actually works. Which is important when solving written equations because everyone has to be on the same page as to what the written expressions actually mean.
great lesson. negative radical stops everything. i got the 'null' as unsolveable.
So, basically, the square root of (2m+2) must equal -5 for the original equation to work. This is what I was thinking… looks like my algebra is quite rusty. Thanks for the explanation. Makes total sense that there is no answer.
This math problem should only exist to explain null answers (or the existence of crappy math problems.)
Forget the square root part, just looking at it and the problem is getting a bigger number than 3 while subtracting from 3. Nothing about it makes sense to my college algebra semester understanding.
There is no way to subtract the positive radical from 3 and get a positive value 8!!
Stars and smilies for you!!
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I would expect the solution is m=(25/2)i - 1, because 2m+2 = 2x((25/2)i-1)+2 = (25i -2)+2 = 25i. With sprt(25i) = sprt (25) x sqrt(i) = 5 * (-1)= -5 the equation is fullfilled.
√(i) ≠ −1. You're confusing it with √(−1) = i
In fact i = cos(π/2) + i sin(π/2), so √(i) = cos(π/4) + i sin(π/4) = 1/√2 + i/√2
There are indeed no complex solutions to this problem.
Complex solutions occur when we have an equation of the form (ax+b)² = negative value
but not when we have an equation of the form √(ax+b) = negative value
² ³ − √ ° ∠ △ × → ~ ≅ π ≈ αβθ ± ≤ ≥ ⁻¹ ⇒ ₁ ₂ ≠ ╪ ·
You've got the identity backwards. You're trying to do √i = -1 but it's √(-1) = i.
That’s something I can’t understand. If someone thinks that sqrt(16)=+/-4 and sqrt(9)=+/-3, what will be the result for sqrt(16)+sqrt(9)?
Will it be either 7 or -7 or 1 or -1?
It looks like there is a solution set.
7 is the ONLY answer
@@user-gr5tx6rd4h Of course, this also my opinion, because I think one can only write sqrt(16)=4 and not +/-4.
IF you defined the sqrt() function that way, then the expression in question has four solutions. The set of solutions is {-7, -1, 1, 7).
@@JacquesLafont There is a workable solution if the domain of the answer is complex numbers or the numerical solution is in the realm of imaginary numbers (sometimes that is true).
For me, what helps me understand what's going on is to hypothesize a simpler example. What is the value of x where √x = -1? Well, let's substitute the imaginary number i so that i² for -1, giving √x = i². Squaring both sides would give x = i^4, or x = i² * i², or x = -1 * -1 = 1. But substituting back into the original equation, √1 does not equal -1, it equals 1. So, there is no solution.
I didn't see how there could be a solution, but since you said to go ahead and use your calculator to get one, I came up with 23/2 and joined the majority.
You have became follower not lead. You have that type presonilty be different do thing on way
Be trend site it sound like don't follow back I recognize the presonilty you want to be I guess word would use heared. Not precisely related I am thinking of trendsiter
You need to go back to kiddy math. SqrtA can not equal -B
No solution. Public school has so failed the majority!
Squaring roots can introduce extranous solutions which is why you always have to check after solving an equation this way.
This is exactly what he is showing here.
@@markmurto That gets into splitting hairs in the real world, particularly in AC electrical situations and in electronics. Half a century ago tunnel diodes were a darling of radio electronics; if biased properly they exhibited *negative* resistance over a small operating range and the speed was effectively at the speed of light. They would still be a major factor in some electronics today if they didn't have a problem of short life (months or less, depending on the service).
In the domain of real numbers (most of the world around us) you are right, In some of the darker paths of exotic tech, you are left behind.
Another way to really emphasize the principle square root as a positive value is using functions. In order to pass the vertical line test, you have to have separate functions, f(x) = sqrt(x) and g(x) = -sqrt(x).
If there is no real solution, then what is the imaginary solution to this equation?
It would be helpful.
Convert -sqrt(2m+2) to i^2*sqrt(2m+2) and solve accordingly. My solution is -1+((25/2)*i^4).
Why should there be one ?
It's been a while since taking algebra and I think I vaguely remember this lesson. I think spending a lot of time with quadratic equations can cause one to forget this more basic lesson. When squaring a number, you have to consider both the positive and negative values that work. When taking the square root, unless otherwise denoted, such as in the quadratic equation with + - square root, it is only the positive root and in this case the positive root doesn't work as a solution and the equation doesn't work as a quadratic.
Isn't the square root of 25 plus or minus 5?
Right!!
no.
Please explain.🙂@@whoff59
The root symbol is for the "principal square root". The the principa square root is a mathematical function which is defined as: "which positive number must be multplied with itself to get the positive number under the root symbol?". This function is only defined for the positive x/positive y-quadrant of the cartesian coordinate system. You as a human know that (2)² and (-2)² both result in 4. But the the principal square root can only reverse the first expression, not the second. That's why you often see ± in front of formulas containing square roots.
no
- SRQT(2m+2) = 5
is the same as:
SRQT(2m+2) = - 5
The square root of any real number is a positive real number.
Therefore the answer is an complex number.
11.5
Wrong answer (try to see if it makes both sides equal in the ORIGINAL equation)
Yes, I was one of those doing algebra in the '60's!
Retired from Power Industry, but now have Parkinson's Disease and was looking for a channel like yours to try to keep sharp!
I saw the null set right away! However, I do not recall having ever heard of the "Principal Square Root" before. Somehow, I learned the concept because -5 for the answer DID NOT occur to me.
I remember that quadratic equations result in the answer being plus or minus, but I don't remember how to check those results. Unless you substitute in the original equation and discard the extraneous result.? Not sure.
I will be sure to follow up with your other episodes!
Semper Fi
Bob
This “teacher” uses a 200 words for a 10 word answer.
God help the poor people he’s trying to teach.
I taught math at a community college for over 20 years and I agree with you completely. This guy sure loves the sound of his own voice.
I totally agree. Public school has failed the majority. Less than ten words. No solution because sqrt(A) can not equal -B. 7 words.
@@markmurto If B > 0.
10-word should be hyphenated.
@@duanehorton4680Help me out, grammar is not my strong suit. Why should there be a hyphen, what is the rule?
I think the main issue is that people seem to remember that when resolving a square root you introduce a "+-" but that's not the case. The "+-" does not come from solving the quare root on the right side but it comes from the left side. x² = 25 means x is a variable and when taking the quare root of x², this is where we introduce the two solutions, not when resolving the sqrt(25). So in some sense it would make more sense to write
x² = 25
sqrt(x²) = sqrt(25)
+-x = sqrt(25)
+-x = 5
=> x1 = +5, x2 = -5
What’s an Algeba step?
Yes, spelling is not his forte. He should stick to the maths.
Yes, I was wondering through RUclips. I found your videos. I’ll try to solve some of your equations most of them right but blew a couple others. I have to tell you that you are an excellent instructor. I struggled with math in high school, l failed algebra, I failed geometry. Sometimes I failed them twice, for my last year and a half of high school I transferred to a different school. I needed to pass geometry which I had failed twice. Praise the Lord for my new instructor it was a whole new game, passed geometry one and geometry two with A’s. I don’t believe my former instructors were bad, but I think Rob knew how to teach me. I tell my friends, who have children who struggle with math I’ to change to a different teacher if possible or find a tutor who clicks with their child. I went to college and luckily had great instructors and TA’s until integral calculus, he was not a good fit. Now days i use my algebra and geometry to make quilts. I plan on watching your videos to add math to my crosswords and puzzles to keep my brain moving. I am 75 years old. Thank you so much.
That is when you hope to step on the green carpet on the sidewalk and find it is just a floating mat of algae. You say "bah!" to complete the meme. Say it!
The Algeba step everyone gets wrong…right after misspelling “Algebra.”
He’s a math whizz not a language teacher 😂😂
@@Redsmeg68
Misspelling stuff is a GOLD mine!!
Look at all the Comments!!!
The radical becomes = -5 so normal root but could equal 5i if you accept complex solutions
5i squared is not 25. It's -25. There is no solution, not even a complex one.
This is great but I'm confused why I couldn't take this path: ... 3 - sqr(25) = 8 ==> -sqr(25) = 5 ==> sqr(25) = -5 ==>(sqr(25))^2 = (-5)^2 ==> 25 = 25
Tidying up, I get -5 = sqrt(2m+2). Since the radical sign implies a positive square root, there is no solution (in real numbers) because the left side is negative. [ Final answer! ] When I check for imaginary roots, I square both sides, and get (-5)^2 = absval(2m+2). Split this up into two possibilities, 25 = 2m+2 and 25 = -(2m+2). This ends up with m1 = 23/2 and m2 = -27/2. Neither of these is equal to 8, so there is no complex solution either. P.S. I also taught high school algebra for 30 years, but I retired 20 years ago, so I'm a little rusty!
When you evaluate a term, you have to resolve a radical before applying a sign to it; there are implicit parentheses around the radical expression. In this example, you can subtract 8 from both sides, giving
-5 - (radical expression) = 0
upon which you can add the radical expression to both sides, giving
-5 = (radical expression),
and you know immediately that won't fly for real numbers.
This is somewhat similar to the kinds of multiple-root fallacies you can get in calculating an internal rate of return (IRR) for a business project. It's a pitfall people too frequently fall into in trying to get numbers to sell a project to management, and, too often, neither those who present the numbers nor the management receiving them understand how such unrealistic predictions occurred until the unrealism comes jome to roost. Moral: don't use IRR to sell something; instead, calculate the range of credible NPVs.
A lot of emphasis there on correct maths, which is fine .. but wouldn't it be easier simply to take the original equation, remind the student that the square root symbol means (by definition, e.g. the requirement to append a +- in the QE formula), the POSITIVE root and so the equation is impossible.
Min 13:38 has little to do with sqrt itself. Rather the fact that x^2=25 x^2-25 = 0 (x+5)(x-5)=0; thus the solution is x=+5 or x=-5. You simply use the fact that sqrt(25)=5 > 0 in the factorization, but the double root comes from the fact that there are 2 factors (x+5) and (x-5).
Video starts at 9:45. You are welcome.
Another way of looking at it is, if you're going to say SQRT25 = ±5, then you can't just choose -5 arbitrarily to make the game come out. Logically, *both* +5 and -5 must satisfy the original equation and this is clearly not the case.
Well, that didn’t go where I expected (but I’ve always considered myself fairly mediocre at algebra). Before playing the video, I just evaluated the thumbnail in my head, started moving things around, and quickly came to the expectation that “the thing people forget“ is that “you’re going to need i” - so I was surprised the answer wasn’t “5i”
I now know why I was so bored doing Mathematics at school, teachers who like the sound of their voice better than giving a concise reasoning in as few words as possible.
Well, the equation has no solution if you stick to a rigid definition of a square root as something always positive, but if you allow yourself to let a negative number also be a square root, it indeed has a solution. What it also shows is that mathematics is not entirely without faults.
No. It just shows you didn't understand square roots.
Apart from that, in mathematics one should always stick to the definition (call it "rigid" if you want) and not allow yourself to let other things be something which they are not.
@@WK-5775 I understand that thing very well. But this implies that I also understand that what is wrong in the first place, is math itself, as it is formulated.
Not a fault, but one of many real world quirks. I would go outside and take a picture to demonstrate, but I can't get through my -8 foot tall door.
I got m = 25i/2 - 1.
3 - sqrt(2m + 2) = 8
sqrt(2m + 2) = -5
sqrt(25i) = -5
2m + 2 = 25i
2m = 25i - 2
m = 25i/2 - 1
Thank you. Great review. I’ll check out your general review of math skills. Took college level algebra and calculus, but never had the same comfort with calculus that I did with algebra, and was probably why I didn’t enjoy physics. Maybe I can move beyond that now that I’ve retired and want to better understand physics beyond mechanics and electricity.
A key thing to notice is the negative in front of the square root after moving the 3. Having that then equal a positive number on the other side of the equation should be an alert to you.
Bingo. Stars and smilies for you!
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Step two is either sqrt(2m+2) = -5 or you can use your step two and make this step three. Once you get to this point, you have a square root equal to a negative number, but square root always means ONLY the positive square root. There is no square root that can equal -5; therefore, there is no solution.
Isn't (-5)*(-5)=25?
And that means the square root for 25 is +/-5.
I was always tought by all teachers that there can be two square roots, a positive and a negative, if not stated explicitly before the square root sign with a plus-sign.
I have never ever heard of "principal square root" and I don't get why it is used here. Still see it is two possible answers with either 23/2 OR no solution.
Maybe this is a new convention that wasn't taught in the 80:s? (or maybe no in Sweden?).
I mean many others writing in here who were taught it decades ago seems to think the same thing.
@@Magnus_Loov It is not new. It has been that way ever since the radical sign was invented in the year 1450.
There is an answer It is 11.5 x i where i is defined as the square root of -1. When I took Algebra, we were taught imaginary numbers when we learned how to solve quadratic equations.
I agree, sqrt(25) = 5, and therefore, there is no solution. 25^(1/2) is not =-5
I can remember from maths lectures during my uni studies of chemistry we should ALWAYS keep in mind that squaring is not an equivalent transformation. Bad things can happen...! 😉
I have a degree in physics. We would use m = 23/2 and the square root of 25 = -5 because that's the one that works.
Would that be spelling Algebra correctly? That was easy!
3 minus -5 = 8
Given that a square root cannot produce a negative result, the solution is null. Or you could just graph both sides; the graph stops as m approaches -3 and never intersects.
Quick question in terms of symbology: does (25)^1/2 mean the same as square root of 25? I don't don't know how to type a square root symbol on my phone but that's what I meant. In other words does using the 1/2 exponent mean both +5 and -5 are solutions because you are not using the square root symbol?
(25)^1/2 is the same thing as √25. So too is 25^1/2 or 25^0.5 or 25^(1/2) or 25(^1/2) or (25)(^1/2), but you'll rarely see two pairs of parentheses. Some viewers use sqrt 25. Some use V25 but that's not very nice. Whether one form or another is used, it's still square root.
All you need to do is to multiply each side of the original equation by (-1) in order to get rid of the negative sign in front of the radical. You will get an identical equation but will not have to deal with the confusing negative sign. Thus, you will get [-3 + radical = - 8], which leads to [radical = - 5], which is impossible for real numbers (unlike for imaginary ones).
It's impossible for imaginary and complex numbers too. There's no value for m, real or complex, that gives √(2m + 2) = -5
Subtract 8 from both sides
U.tiply -1 on both sides
Results. 5 + sqroot (2m+2) = 0
Retake 10th grade. Sqrt [a] can not equal -b. No solution by simple observation.
OK, since 5 is already greater than 0, you can see there is nothing you can add to it to bring the answer down to 0, since SQRT(X) cannot be < 0.
If you are stuck with the idea that the solution set has to include both 5 and -5 for the square root of 25, keep in mind that the reason quadratic equations result in a solution set of two possible values for the variable is because _both_ values will result in a correct solution. In this case, one of the two possible values does _not_ result in a valid solution. This means that there is no valid solution set. And yeah, I was stuck m = 11.5 too.
The spelling step that this channel got wrong. ALGEBA for heaven's sake.
You've essentially demonstrated that the solution the majority of people will reach is correct.
I challenge you to demonstrate how this has been a problem in industry.
Mr michael intelligent people dont boast. A high school learner will be able to give the answer
Just a sidenote:
At 12:37, the resolution to the equation shown at the right of the screen yields the correct result but there's a problem on the second step.
We know that √x²=|x|.
Yet you wrote √x²=±√25 which is wrong.
It should be: √x²=√25
Then |x|=5
And finally x=±5.
By saying that √x²=±√25 you would be saying that √x² can be a negative number, which it cannot be.
Just a small inaccuracy that somewhat contradicts the main point of the video.
Now, that resolution to the equation may very well be there to demonstrate how some people may get that step wrong while solving for x, but as it is not stated anywhere on the video i suppose it's just a small mistake that made it's way into the video.
You made the mistake you are trying to avoid at 13:24.
"So the square root of x squared is x" - that is not true. It is important to not reinforce this fallacy.
sqrt(x^2)=|x| not x.
|x|=5 => x=5, x>0 and -x=5, x
You tell him.
Once you get that the square root is a negative then you could only get a solution if you use complex numbers i.e. i in other words 2m + 2 = sqrt(5) * i Without using complex numbers there is no value that you can square to get a negative number.
I simplified question to 3-x=8 Seeing x had to be a negative and intuitively knowing you can't get a negative from square root knew something was wrong. First thought it might be imaginary but that didn't work out either so knew there was no solution. Did I just get lucky or was logic correct..
This content was a fantastic addition to my learning!
My algebra is a little rusty after 40 years, so I tried to look at this problem logically - not that algebra isn’t logical…..lol😂
Fundamentally, 3 minus anything cannot = 8 unless you’re subtracting minus 5 but Sq.Rt of (2m + 2) cannot be minus 5. It’s basically the same logic.
An introduction to *_Social Justice_* mathematics:
a = b initial supposition
a*a = a*b multiply both sides by a
a*a - b*b = a*b - b*b subtract b*b from both sides
a*a - a*b + a*b - b*b = a*b - b*b substract and add a*b simultaneously on the left side
a*(a-b) + b*(a-b) = b*(a-b) factor out (a-b) thrice
(a+b)*(a-b) = b*(a-b) combine a+b
a+b = b divide out (a-b)
b+b = b use the first equation a=b
2*b = b reduce b+b to 2*b
2 = 1 divide out b
3 = 2 add 1 to both sides
4 = 3 add 1 to both sides again
5 = 4 ... and continue adding 1 to both sides
1 = 0 subtract 1 from both sides of 2 = 1
0 = -1 subtract 1 from both sides
-1 = -2 subtract 1 from both sides again
-2 = -3
-3 = -4
-4 = -5 ...
... -5 = -4 = -3 = -2 = -1 = 0 = 1 = 2 = 3 = 4 = 5... and so I have proved that all integers are equal!!
But some integers are MORE equal.
When I was in school, back in the ice age, we called the 0 (null) answer as "undefined''; does that answer still apply today?
"undefined" and "no solution" are the correct ways to express it. What he wrote ("m = ∅") is incorrect. If you wanted to use set notation, you'd have to write "{m} = ∅".
Mathematics needs a symbol for the operation that returns a positive and a negative value for the square root, in other words for the reverse of the square function. There needs to be an operation where the value for -5 (or some other number) squared can then be reversed to return the original value.
We have a symbol if you want to refer to both the square roots: it's ±√25.
Thanks for this great video !!!
The first step is move the 3 to the other side - (2m + 2)^1/2 = 8 - 3. square both sides and one gets
2m + 2 = (5)^2. The final solution is m = 23/2
The problem arises in "square both sides." For example, take the equation 10 = -10. Square both sides: 100=100.
This was informative. Thanks!
It’s a miracle that we made it to the space without your help, bro
4:56 You can't square the negative...
You should divide by -1 first - and then you have SQRT(2m+2)=-25 - which isn't possible.
What I noticed right off the bat is that √(2m+2) has to be -5. A real square root can't be negative.
The result of the square root of 25 is the value of 5. If you square a positive 5 or a negative 5 you still get the same result of positive 25. That is why the square root of a number technically is +/- the magnitude of the result of the square root operation.
@@GaryBricaultLive Yes, I got confused. The square root of a number can be negative. I was thinking of the square root of a negative number, which is a different thing.
@@GaryBricaultLive Your first and last sentences contradict each other.
Not even complex numbers can have negative principal roots.
@@wernerviehhauser94 There is sometimes value in the negative aspect of a square root. If we are going to fall short, it may be helpful to know whether we can come up winners next time. It won't work this time, but may define a fix next time.
2m=1 I think.
9^1/2 - (2m+2)^1/2 =8
-2m+9-2=8 | -2m=8-7 | -2m=1
2m=1 m=-1/2
Now let's look at what you did lol na I screwed up combining the roots impossibly
I have an intuition that the +5 solution would draw a different shape geometrically than the -5 solution. I have no intuition that sqrt(25) does not include -5 as an answer.
Hey this is interesting. Without having watched the videos, here's some thoughts.
√(2m + 2) = -5 can't be solved on the real plane, so it must be a complex something. It looks like the modulus of a complex number [Z_modulus = √(a^2 + b^2)]
Z_modulus = √(2m + 2) = -5
Write that as the conjugate [Z_conjugate = a - ib] and you get the following I guess?
Z_conjugate = √(2m) - i√2 = ???
Don't really know how to continue, or if these even remotely made sense or is useful.
When I invented Algebra back in ‘68 I didn’t realize how confusing it could get. If I could go back and do it again I would leave out all the letters and just stick with numbers.
Algeba was invented WAAAY before '68.
Wow, really wonderful observation there. You mean to tell me that I didn’t actually invent Algebra? You mean it was actually developed over 3000 years ago? Thank goodness there are sharp people like you around to make sure jokes aren’t taken literally. I’m sure all of the people who thought a random RUclips commentator invented algebra are grateful for your wisdom.
Thanks; What if the positive sqrt gives the correct answer, but the negative sqrt does not (i.e. if 3-sqrt(2m+2) = -2 ) Do I also conclude there is no solution?
Mr. Math Man!
Mr. Math Man!
Mr. Math Man!
That the sqr is only positive is only an arbitrary definition, not something fundamental. In the real world you should take it as a possibility that the writer intended +- sqr. Especially if they state that there is a solution.
sorry to break this to you, but this has nothing to do with "real world applications" and all with "didn't really understand how numbers work". If the author of the question intended a +- square root, he was in error.
Yes and no. The only time I used my advanced math skills in the real world was to calculate the value of a series capacitor I had to put in line with a wire antenna to bring the capacitance down to spec. If the value had been imaginary I could have looked at shortening the wire antenna because it would not work this time.
The algeba step everyone gets wrong - missing out the "r" in "algebra" :-)
I commented so much but this person is still explaining
Title should be: "The Spelling Step that WE Got Wrong!"