The Algebra Step that EVERYONE Gets WRONG!
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- Опубликовано: 7 авг 2024
- How to solve radical equations correctly.
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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.
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
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.
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.
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.
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
This was informative. Thanks!
Great,many thanks !
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
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 video was extremely helpful, thanks bro
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.
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.
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.
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.
that was really informative, if you rewrote it as 3-(2m+2)^(1/2) = 8 would there be a solution? i'm guessing no
NO because the square root encases the 2m+2 as follows [(2m+2)^(1/2)]. As usually this guy's problems are poorly defined with lots of ambiguous aspects.
I write the square root of x^2 as |x|=5
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.
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"?
Writing this before watching the video so I can have my thought process out there and see how it compares.
3 - sqrt(2m + 2) = 8. Subtract 3.
- sqrt(2m + 2) = 5. Multiply by -1.
sqrt(2m + 2) = -5. Square to get rid of sqrt.
2m + 2 = (-5)² = 25. Subtract 2.
2m = 23. Divide by 2.
m = 23/2 = 11.5
Got the error directly: the r is missing!
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.
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.
Just curious why you
didn't use PEMDAS this time. Because then you could have dealt with the sqrt first. I am not an expert so any response is acceptable.
The given expression is NOT an equation, hence to look for a value of m which would satisfy this NON-EQUATION would be an exercise in futility.
Expressions are not equations. What he wrote is an equation (not an expression). The "=" sign is a dead giveaway!
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).
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} = ∅".
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).
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.
Excellent
i'm an adult that's brushing up on my math skills. i find your classes interesting and helpful. thanks.
Lol keep it up you’re helping me keep the brain fog away TY
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.
This can’t be solved with real numbers. But it is easily solved with imaginary numbers.
Set α= 2m+2. We want to solve √α = -5. In complex numbers that’s trivial. α= ± 25 i where i is the square root of -1.
Then we solve 2m+2=± 25 i. 2m= -2± 25 i. m=-1+ 12.5 i
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.
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
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.
If you manipulate the equation to get equivalent of y=(-sqrt(2x+2)+5 and graph it on Desmos, you indeed get only one point on the x axis: 23/2.
How do you show that you want the negative square root? There is a standardized way to do so right?
Put a minus sign in front of it. Or put a +/- sign in front of it if you want both.
@@davidbroadfoot1864 thanks
That's amazing handwriting for paint
3 - square root of (2m + 2) = 8 subtract 3 from both sides
square root of (2m + 2) = 5 square both sides
2m + 2 = 25 subtract 2 from both sides
2m = 23 divide both sides by 2
m = 23/2 = 11 1/2 Is that it ? I'll check now.
Square BOTH sides
@@christopherellis2663 Yes .. square BOTH sides of the equation
The square root of (2m + 2) SQUARED is 2m + 2
and .. on the other side of the equation .. 5 SQUARED is 25
3 - square root of (2m + 2) = 8 subtract 3 from both sides
square root of (2m + 2) = 5 square both sides
This unfortunately is wrong. You get
- square root of (2m + 2) = 5 square both sides
Don't simply lose the '-'!!!
There's other ways of becoming famous
Great video! LOL, you made me feel bad and then good again:-)
I could simply not find a solution, but I thought there had to be one, so I tried and tried again. When you then answered that there were no solution, hen did I feel good:-)
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.
How would I use this in the real world? What kind of problem would I encounter that would require me to use this skill?
I can't give you a specific real-world example, but the same could be said of most puzzles. Flexing the brain builds your problem solving skills, which has innumerable real-world applications.
@@aek03030731 Thank you for your kind reply. I am seventy years old, and I am trying to relearn mathematics, particularly Algebra. I often wondered while I attended college was some of the mathematics required, just used as barriers to entry for some professions to keep numbers small. Also does requiring additional mathematics provide jobs for professors? I understand some Medical Schools require additional calculus for admission. The additional calculus courses will never be used by the Medical Doctors, just used as a barrier to entry to Medical School. I worked in industry. I noticed the difference in engineers who went to colleges that taught engineering as theory and engineering as application. I also noticed that engineers that went to "good" engineering schools were not necessarily good engineers. They might been intelligent, and proficient at mathematics, but that was the extent of their engineering abilities. They did not want to move from their desk. Some guys with a two year degree from a community college, were better electrical engineers than those that had a Bachelor's and even Masters degrees in electrical engineering. With the high cost of college, I have talked to a few recent college graduates who ask the same questions. Is all the mathematics just a barrier to entry into a profession? Once on the job they never use all the math that was required. Again that you for your kind and honest answer.
This equation is an inverse function of a quadratic whose y intercept is 23/2.
Couldn't the solution be m=(23/2)i where i squared = -1 and is an imaginary number?
You didn't do what he said you should do, which is to plug it into the equation and see for yourself. Then you will see your error.
no. Even imaginary or complex numbers can not have a pricipal root of -5.
What Is the answer?
There is no solution.
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.
@@user-gd9vc3wq2h 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.
great lesson. negative radical stops everything. i got the 'null' as unsolveable.
Umm ...is it spelling _Algebra_ without an 'r'?
That it?
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.
I assume you are dealing only with real numbers. Thus sqrt(2m+2) if exists is >=0 sot So 3-sqrt(2m+2)
Does anyone know if the “Principle Square Root” convention is solely an American thing, or is it employed in the UK and other developed countries as well?
It’s just that it seems crazy to me. There doesn’t seem to be any logical reason for it.
The Principal Square Root idea is good because it lets you quickly see that there is no solution. if you don't like that idea, its fine; You can do the problem the hard way - that is, search for a value of M that makes the equation work - and there is no value that works. you can't use 23/2, or -23/2, or any other value.
The principle square root is *a* solution, it is not the operation.
The only reasons I can think of for it are computers or lazy instructors. Mathematical functions can only return one value (one value of y for any value of x for y=func(x)). Square roots are an operation (not technically a function) and actually actually usually have two solutions. A function can only return one, so the function returns the positive solution, or zero (or an error if x
@@philip5330. Wrong. Plug in 23/2 for m. It works. But only if you accept that the square root of 25 can equal -5 as well as +5.
@@pmw3839 It's all about notation. The question reads as "3 minus the principal square root of (2m plus 2) equals 8". So although -5 is a square root of 25, it doesn't apply to this equation.
@@davidellis1079 How are we supposed to know that? I mean you say yourself that -5 is a square root of 25. I see no explanation for why we should not use it here!
My process:
3-√(2m+2)=8
3=8+√(2m+2)
3-8=√(2m+2)
√(2m+2)=-5
2m+2=(-5)²
2m+2=25
2m=23
m=23/2,
which tells me the equation was not seeking the psqr of 2m+2.
Why am I wrong?
√(2m+2) cannot equal -5
No square root can have a real solution that is negative.
@@jamesholden4571why do we need a real solution why can't it be i? Regardless the question is asking what does m equal not to get hung up on gibberish of square roots.
You are not wrong. M does equal 11.5. We have our answer and can move on. The debate people are waging over principal square roots is moot. We have two possible solutions for the square root. Use whatever one gives m = 11.5. Your process ignores the messiness of complex numbers so is the correct way to approach the problem. Other methods are undesirable as it confuses people and they get lost in semantics.
There is something wrong with the way this problem is written. Therefore it is not an application requiring a real world solution but an exercise to test if students can correctly isolate and solve for one variable. They were trying to trick you and failed. Congratulations you get an A.
"When you use -5 you get the right answer"... he said that! Then said it was wrong. Change the rules to suit the problem.. how do you then justify (-5)^2=25??
I agree, sqrt(25) = 5, and therefore, there is no solution. 25^(1/2) is not =-5
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.
Er, how do you spell algebRa?
It is the spelling "everyone gets wrong!" - allegedly
What's algeba?
Shouldn’t the correct answer be 59.5i ?
can we not use imaginary numbers?
That won't help. There's still no solution.
The genius is back!
m=12.5i^4 - 1 where i^2=-1
Thank you! It seems the teacher forgot the step of telling us the answer had to be a real number!
I do not understand why the principle square root was mentioned here and the problem was not solved. Either you mention it and solve the problem or do not mention it.
He did solve the problem. The answer is that there is no solution.
If we define the PSR as always >= 0 then the problem itself is false, like saying 3=2
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?
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.
11.5
Wrong answer (try to see if it makes both sides equal in the ORIGINAL equation)
So you get to have a principal square root. that buys me the ability to have a principal answer 13/2. the problem is the negative sign in front not that the fact that 25 has a positive square root. Time spent with numbers means almost nothing. I'm sure that Gauss and Euler both were exceptional as pre-teens.
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...! 😉
The algeba step everyone gets wrong - missing out the "r" in "algebra" :-)
If one only uses the principle square root, the entire concept of solving quadratics by completing the square becomes unusable.
no it doesn't
it works perfectly fine with the principal root. If you think it does not, they you either had a bad math teacher or were a bad math student, since there is obviously something off with your understanding of the problem.
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.
I commented so much but this person is still explaining
Algeba?
Start with 120 cm……..first step is 1/3 of total ( 40 cm ) …….every next step is 1/3 the distance of the previous step……..how many steps to pass 60 cm mark ?
Good math texts should make a big deal of teaching the principle square root so that when a student studies quadratic equations they won't default to this misunderstanding. This is an important issue in teaching math.👍🏾
I'm an electrical & electronic engineer & I have to work with ALL the roots & CANNOT simply just take the singular "Principle Square Root" as they all give solutions that need to be evaluated... just as if I take the 4th root of 16 it is 2, - 2, 2i & - 2i...... if I answered that the 4th root of 16 was ONLY 2 I wouldn't get many points. Just like cube root 8 has one real (=2) AND two complex solutions ...... & before you answer that you simply have to use +- if you want to designate both roots that only works for square roots. +- doesn't work for higher order roots which go complex. The only way it works is if the root sign means all the roots not just the principal one.
Interesting question if 2i to the power 4 = 16 then what number to the power 4 is 16? You are saying 2i is wrong & not a valid solution when it is.
I agree with the idea that if we consider complex numbers, one can say that 4 has two square roots, 2 and -2, which are real numbers, and -4 has two square roots, 2i and -2i, which are not real. But for me, one can use the symbol V’’’’’’’’ (I can’t write it better!) only for the positive square root of a positive real number. So when one write sqr(4)+sqr(9) with the symbol V’’’’’’’, that means only 2+3=5. (and I don’t use this symbol for the complex square roots of -4)
When you are in school or college, you are supposed to do what the teacher, instructor, professor, etc. tells you to do. What he or she tells you to do is most likely related to the principles of whatever subject you are studying. So, if you are studying algebra, then you have to consider principal roots sometimes. If you are studying electrical engineering, then you probably have to consider all possible roots. If you are working for an engineering firm, then you probably have to consider all the roots since the aspects of your job might require that.
Here's a baseball analogy. When your baseball team plays in your home ballpark, then you are supposed to follow all the ground rules that concern your ballpark such as what counts as a ground rule double, is it a home run when the ball hits the yellow stripe or is it still in play when the ball hits that same stripe, etc. If your team is playing at another ballpark, then you have to follow the ground rules of that ballpark.
There is a difference of
a) 4th root of 16
and
b)solution of: x⁴ = 16
a) is only meaning +2
b) has 4 solutions +2, -2, +2i and -2i
sqrt(2m+2) can not be equal to minus 5.
No solution.
Check the main title
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.
11.5i
m is complex number which makes it solution
There's no complex value for m that satisfies this.
m is - 13,5 2x (-13,5) + 2= - 25 -sqrt of -25 is 5
"-sqrt of -25" is not 5 ... not true. It is -5i.
Algeba? I believe you may be missing an elbow. Or, is that an r
Doin Algeba on Yubatuba
Sad to say I would be happier if this was solved in the complex plane with sqrt((-1) = i
In the past I could have done it this way, but not now.
Roots are (-2+ root 5)/2
Might want to check your spelling on the opening screen! Love your videos.
Hes a math teacher, the only letters he cares about are x,y, a,b,c ,d, t, v, u 🤣
there is a much easier way to explain this. The algebra we learn to do has a hidden assumption in its methodology which is that the steps must be reversible. This is why you plug in the answer to check. extraneous solutions are asymmetrical which is why you cant use them.
C+ sounds a bit mean to me, just for getting one small, rather obscure point wrong! I'd have given a B and saved the A for a correct answer.
well, getting small points wrong in medicine and engineering kills people....
C+ is very generous. My teacher would have maxed out a D, and so do I.
If everyone gets it wrong maybe you should do something about it as its obviously not being taught correctly. I can tell you 100% i have never heard the term principle sqrt ever when learning about sqrt.
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.
It is a CONVENTION that when solving for square roots, only the principal square root is used. However, the secondary square root still exists. So think otherwise is as foolish as believing that negative square roots don’t exist simply because they are referred to as “imaginary,” and do not show up on the Cartesian graph of real numbers. But imaginary roots still exist, and you ignore them at your peril.
If you follow the principal root rule, then there is no solution. If you were to put a plus sign in front of the radical instead of a minus sign, then you would have a solution.
.... like spelling algebra "algeba" on the blackboard?
-5
WHAT ABOUT THE SOLUTION THAT THE BOARD OF EDUCATION REVIEWED THIS AND FIRED YOU?
5i
0.707