The square root of (3x) = - 7 Don’t make this common algebra error!
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- Опубликовано: 30 май 2024
- How to solve a radical equation and check for extraneous solutions - null, empty set. Learn more math at TCMathAcademy.com/.
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As a 73 years old grandmother, I declare that WE CERTAINLY NEED MORE MATH TEACHERS LIKE
MR. JOHN, WHOSE SERVICE IS PRICELESS, ESPECIALLY FOR THE YOUNGER GENERATIONS 🙏🏽
I wish that I had this kind of opportunity to relax and learn math via a RUclips channel, run by a dedicated teacher, when I was a student.
This gives extra time without any stress to understand these nitty-gritty details in math!
This also demonstrates a positive outcome of the advancement of technology!
You should adopt ‘math with Mr J’ method. Brief and clear..
You have to put the ± in front of the √ in order to look for a negative square root. That doesn't mean that the negative square root number is wrong, just that you were not looking for it. So the equation would have to be ±√3x = -7 for there to be an ACCEPTABLE answer.
That was very helpful. Thank you.
You're correct (of course) but, when pushed, my calculator came up with: 1.7320508*sq.root of x = -7 (ie: None of the Above)!!
You got me. I did it fast and got a. I totally forgot about.... Now listening to remember.
Pretty neat math teaching.
Got me on that one. I was sure that the answer was 49/3. I learned something here.
How sweet it is when we can acknowledge like that with gratitude!
How about x = ( 49i⁴) / 3
So the 3's cancel
The square root of 49 is 7.
( i⁴ )^(½) = i ^ ( 4 • ½ ) = i², then
i² = √-1 • √-1 = -1, so
-1 • 7 = -7
Do you think that works?
No, because √(i⁴) is not i².
√(i⁴) is -i²
Why?
@@stucampbell1274 i⁴ is just a fancy way of writing 1, so √(i⁴) means √1.
1 is a positive real number, and √(any positive real number) is positive. That's the definition of the √ symbol. √1 = 1
i² is just a fancy way of writing -1. So √1 is not i², it's -i².
Okay, so I figured there was something funny going on here when I saw a negative sign and a radical. Those two things are like ammonia and chlorine--unless you using wearing protective gear in the form of imaginary numbers, you want them as far as part as possible.
Great analogy.
Imaginary numbers don't help here.
I have a very simple philosophy. If I have to solve a difficult equation and I'm not confident of my answer, I substitute it back into the equation. He says the solution is x=0. So cam someone explain why the square root of zero is -7. The only one of the 4 answers offered which works when substituted back into the original equation is, of course, answer a). I think this solution is seriously flawed
@@philipsamways562 Where does he say that x=0? That's definitely not correct.
Answer a) isn't correct either. That would give you
√(3x) = √(3 × (49/3)) = √49 = 7
But we're not looking for 7, we're looking for -7.
There is no solution to this equation.
@@philipsamways562 That crossed-out circle is not a zero. It's the symbol for "no solution". And there isn't, not even in the imaginary plane. Believe me, I tried.
Thank you
This channel is for those who are starting to learn and to understand the principles and in some for review and fun.
If you are already a genious in math get your own channel.
But can I solve it like this
V3•Vx=-7
Vx=-7/V3
(Vx)^2=(-7/V3)^2
x=+,- 49/3
Explain why not.
Thanks in advance.
When you square -7/√3 you don't get ±49/3, you get 49/3. Your solution doesn't give x = ±49/3. It gives x = 49/3.
But when you substitute x=49/3 back into the original equation you end up with
7 = -7, therefore x = 49/3 can't be the solution.
So why doesn't x=49/3 work as the solution? Because the original equation isn't valid. The original equation says √(3x) is negative, but √(anything) cannot be negative by definition.
The squareroot of any number can be +ve or -ve. For example, sqrt(49) can be +7 or -7.
No it can't. He just discussed that. When taking a square root, only the positive is correct. The negative is extraneous.
It's true that 49 has two square roots. But the expression √49 refers to exactly one of those square roots: the positive one. That's the definition of the √ symbol.
If you want to refer to the other square root of 49, i.e. the negative one, you have to write -√49
@@gavindeane3670 : Exactly. You only can about both when it's of the form √(x²). Besides, people need to check their answers by plugging them back into the original equation. Now, if they're looking for the solutions of a quadratic, then yes, there are two solutions.
@@argonwheatbelly637 √(x²) doesn't refer to both square roots. √(x²) has a single value. It equals the absolute value of x.
If you want both square roots, you have to write ± before the √. The formula for solving a quadratic is a perfect example of that.
Correct.
Yeah Mr John don't you ever listen to that garbage. Please I need you to take the time to show me. I never knew there was a thing like that that meant it has no answer. So I'm learning and I know that's what you care about. It's your passion.
People like that, thank God for them it wasn't an IQ test. 🤣
I'm gonna give you onna these 👍 and 👋💪❤️🌎
I have a serious math problem 😂
Thanks again
You didn't even metion imagery numbers. How hard a concept is i aka the square root of negative one?
There's no reason to mention i when discussing this question.
Well I thought the symbol was zero. I forgot about null symbol. So when I got to 7=-7 that wasn’t one of the multiple answers. So I watched to find out what I did wrong. Smacked myself when you said null the first time. I was like the memory was jogged and I thought…No s(85 Sherlock. I’m 65 haven’t done this since I was 17 so really just forgot about null set. Oh well. That’s why I watch you. To try to keep my math skills up.
Holy crap that's amazing thank you ❤👍👋💪🌎
So in love with himself
X to power 1/2 is a second degree equation so 7 and - 7 are the answers.
It is true that 49 has two square roots: they are 7 and -7.
But there is no solution to this equation because the √ symbol refers to exactly one of the square roots: the positive one.
√(anything) cannot be negative by definition.
No. X to the power ½ is just a different way of writing the Square Root of X.
It evaluates to only a single value.
You can't sidestep the issue by rewriting the radical as a fractional power.
Please do your introduction and your appeal for subscriptions if need be at the very beginning ie before you work on the problem. This can avoid any distractction
Come to the point quickly, cut out the meandering drivel
Is it a
negative 7 is a REAL number!!!!! you need to write it as the real positive number R+
a) 49/3
Correct! Every non zero number has two square roots and every proper equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
d. in about five seconds in my head. Again, people, you need to check your answers!
got it D no negative result possible thanks for the fun
A)49/3
Correct!
Every non-zero number has two square roots and every equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
49/3
Every non zero number has two square roots and every equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
The square of -7 is 49 (positive), so it should be fair to argue that BOTH 7 and -7 are square roots of 49. This is just another example of where math itself gets weird. Not sure if sometimes math concepts don't correspond to the real world, or if human language is inadequate to properly define the math concepts.
correct, but ONLY if you are talking about x². We are NOT using x² here so there for we can not use that rule.
There are no solutions, not even in the Complex world.
Answer is a) 49/3
Correct. Every number has two square roots and every equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
unsolvable
A___49/3
Correct!
Every non-zero number has two square roots and every proper equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
No need to get into the bother of trying to solve it, or checking for extraneous solutions. We can immediately see that this is an impossible equation because √(anything) cannot be a negative number.
Yes it can. Why can’t it? The imaginary number (i) is only necessary when you are taking the square root of a NEGATIVE NUMBER. In this case, you are taking the square root of a POSITIVE number. That doesn’t eliminate the ability to have negative square roots
@@matthewcastleton2263 Of course numbers can have negative square roots. The two square roots of 49 are 7 and -7.
But the definition of the √ symbol is "principal square root". It refers to exactly one of the square roots of its operand - the positive square root.
If x is a positive number then the two square roots of x are √x and -√x. √x is a positive number. -√x is a negative number.
If you want to refer to both square roots - see, for example, the formula for solving a quadratic equation - you write ±√
√49 = 7
-√49 = -7
±√49 = ±7
@@matthewcastleton2263 Only if it were quadratic.
@@matthewcastleton2263square root of a real number is defined to be the positive branch. Just graph sqrt(x) and see that.
@@matthewcastleton2263 There is no Complex number of the form x=a+bi such that when you take the Square Root of x, the result is a negative number. No Solution, not even in the Complex world.
x=(49i^4)/3
If that's the value of x then 3x is a positive real number, therefore √(3x) isn't negative
i⁴=1 so that goes back to x =49/3
Correct!
Every non-zero number has two square roots and every proper equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
...hhmmm...The principal square root cannot be a negative (-) number! The question is NOT possible, no way, no how!
1 and 1 is 2😮
Basically it is 1.999999999 but you can round it to 2 😅
I am happy for that explanation. I thought the answer was 49/3.
It is 49/3. This fool is trying to pull the wool over your eyes
That's the point of this question. He posts an equation like this from time to time so that he can explain the exact meaning of the √ symbol, which isn't always properly understood.
@juniorleslie4804 Same here. :(
49/3 IS the correct answer.
Every number has two square roots and every equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
d)
a
I came up empty. Because a square root can’t be a negative so I will take 0
Yes it can. You just can’t take the square root of NEGATIVE NUMBERS. This fool is GASLIGHTING. The square root of a negative number is the only one that requires the imaginary number i. You absolutely can get negative square roots because two negative numbers multiplied by each other equals a positive number
The solution is 16 1/3 i
Where did you get that from???
If that's the value for x then √(3x) would be √(16i).
√(16i) is nothing like -7.
There is no solution, real or complex, for the equation in the video.
@@gavindeane3670 You are wrong.
Every non-zero number has two square roots and every proper equation has, at least, one solution.
In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
CORRECT!
Every non-zero number has two square roots and every proper equation has, at least, one solution.
In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
@@davidbrown8763I'm not wrong. Yes, numbers have two square roots, but the expression √a does not mean "either or both square roots of a, whichever you happen to find convenient". The radical symbol is more precisely defined than that. √a means, specifically, "principal square root of a". √49 equals 7 and only 7. That is the definition of the radical symbol.
If you want to refer to the other square root it's -√a. If you want to refer to both square roots it's ±√a.
√49 = 7
-√49 = -7
±√49 = ±7
x = 49/3 is indeed the solution (the only solution) to the equation
-√(3x) = -7
but that is not the equation we're being asked to solve here. The equation we're given has no solution.
@@gavindeane3670 I understand where you are coming from, but most people are not that pedantic, and the chances are that most examiners also assume what I have stated. Every Maths text book I have seen assumes that all non-zero numbers have 2 square roots and every proper equation can be solved. (An example of an improper equation would be one that is obviously silly, such as 3x +7 = 3x - 7. Here the 3x is eliminated,, leaving 7 = -7, because you cannot add something to anything and expect is to be equal to what you subtract from it. In other words, it is clearly not an equation).
Hence the chances are that you would be marked incorrect - especially in a multiple choice question (which this is), because you do not have the opportunity to explain why you chose d. So I would still take my chances on choosing a..
You should note that the imaginary number i is required to solve this problem.
The imaginary number i doesn't help.
@gavindeane3670 You must not have a good imagination. The number i is just as real as Santa Claus, The Easter Bunny, and The Tooth Fairy, but Krampus is made up.
I came looking for that as none of the answers in the video preview worked.
No, there are also no Complex solutions for this equation.
With all respect, I do not agree.
Every non-zero number has two square roots and every proper equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
I appreciate your concern for explanation, but it kinda distracts from the problem at hand creating a lost of interest , so kindly shorten the narrative a little .
-21
🙂
D
B
So, the actual solution is something like 49i/3, a complexe number.
No it’s not. It’s 49/3
No. There is still no solution even if you consider complex numbers.
Complex solution gives you 49i⁴/3 which resolves to 49/3
@@martinkeepin394349i⁴/3 is not a complex number, as you have noted. It's also not a solution to this equation.
@@gavindeane3670 well, written in that form then it's complex but yes, it resolves to a real number
Even calling this, the square root is an error, which should not happen in the math class. Square roots can be negative, but radicals cannot. This is the radical of 3X, and the radical can never be negative by definition. It is misleading the student to call this the square root. This is not good math teaching.
Finally, the proper response. Thank you @rodrodrigues5402
It's "The Square Root." (singular), or more properly, "The Principal Square Root."
In either case it means the non-negative number that when squared gives the indicated value.
Pls shorten your video!
Plz don't 🤪
Like A LOT
He won't, I think he loves to a) hear himself talk and b) loves restating everything and dragging it out.
@@edwardharrelson6988 smart answer. Give yourself a big smile and an A+++ 😜
@@edwardharrelson6988 and some of us (me) prefer quality explanations, youtube is filled with "tutors" who just show off hardly explaininv anything
STOP GASLIGHTING YOUR STUDENTS!!!! You simply asked for the answer to the solution. Negative 7 IS the square root of 49!!!!! Yes, 7 is the PRINCIPLE square root of 49, but that doesn’t make the answer of 49/3 incorrect. Because, without a doubt, negative 7 squared most definitely is 49, making negative 7 a possible answer for 49^1/2. Unless you specifically specify in this equation that ONLY the principle square root is acceptable, then you are GASLIGHTING
He did explicitly specify that only the principal square root is acceptable. That is the definition of the √ symbol.
@@gavindeane3670 but not in the original equation. If that is how the equation was written on a test, with no qualifiers stating that only the principle root was acceptable, that would be a misleading question intentionally crafted so the ones reading it get it wrong. It only states to solve for x. That’s intentionally misleading students so they will get the answer wrong on the test even when they solve it in the proper method, making them think they are crazy or doing something wrong. It’s gaslighting
@matthewcastleton2263 Not sure what's happened to the reply I posted here, but I think we already covered it in replies elsewhere.
In summary: the equation in the video DOES explicitly state that only the principal square root is being referred to. That is the definition of the √ symbol.
Too many words
Before watching. Sqrt of any real number (times 3 , or not), equals a negative number? The answer is no solution in the domain of real numbers. Is this video going to get complex? Continuing to watch now.. aha, it comes down to the test style, multiple choice. In this case it’s easy, I assume 8th? Grade. In 12th grade I would expect some student to be a smarty pie and actually solve it in C🤓 how would you grade an 8th grader who gives the complex solution? Take them to the principal’s office, threaten them with expulsion, as a rather cruel joke, and then give them a subscription to some advanced math, or physics, resources. Happened to me and 2 others in 8th grade on a different topic. I still don’t know if I am appreciative or upset at that moment.
There's no solution at all, real or complex.
I concede . Did not try to solve it myself before my comment. I checked now and , yes, this has no solution, even in the outer realms 😔 going metaphysical, one exception, in a galaxy, far away , and a long time ago. Define a new algebra that contains this specific equation as a solution as an axiom. But now we have left reality for sure 🎉
a) 49/3
Correct.
Every non-zero number has two square roots and every proper equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
I am glad that Mr John, explained well that the correct answer is not 49/3.
Yes, if it is the correct answer, it should produce -7, when it is plugged into the original equation.
But, when x= 49/3 is plugged in, it produces the principle square root of 49, which is only +7.
So, 49/3 is not the exact accurate answer and it could trick us, if we forget about this trap in mathematics!
So, we have to declare that the answer to this problem is invalid and it is an “extraneous” solution, which means that it does not yield
the correct answer when plugged in to the original equation.
a
a
Every non zero number has two square roots and every equation has, at least, one solution. In this case x = 49/3 is the solution to - sq root 3x = -7 and + sq root 3x = -7 is extraneous.
a