Why don't these cancel out? The square root of x^2 is not always x!

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When teaching Calculus last semester, this one algebra concept proved to be the trickiest for a lot of students. The square root of x^2 is just x, right? There's a subtlety that can trip up a lot of people. Let's look into this common misconception and see how to fix it!

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25 янв 2024

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Комментарии : 25

@malemsana_only 6 месяцев назад

Brother, keep making videos.

@DrSeanGroathouse 6 месяцев назад

Thanks! More videos coming soon

@Blackmuhahah 6 месяцев назад

The arctan/atan2 functions are something I encounter quite often

@andrewfoley 6 месяцев назад

Best Calc teacher ever. Thank you for a great semester. 😄

@ShaolinMonkster 5 месяцев назад

I've seen all your videos. Perfect size and interesting topics, well described. Good work!

@DrSeanGroathouse 5 месяцев назад

Thanks so much!

@atrus3823 2 месяца назад

The sqrt of the square of a variable being equal to the absolute value of x was hugely resolved sooo many issues for me in my time taking math in school. I’m surprised it’s not taught more. I think this should be a base identity you learn in like grade 10 or something.

@bhgtree 5 месяцев назад

Perfect explanation, I've watched two of your videos and you've done the best and clearest explanations I've seen on YT, Thanks.

@DrSeanGroathouse 5 месяцев назад

I'm glad to hear that!

@strigibird9832 5 месяцев назад

Wouldn't utilising imaginary numbers solve the problem with square roots of squares? Let's say we write -3 as 3*i^(4n+2). Squaring that number would bring you 9*i^(8n+4) which does indeed result in a positive number as the power i is raised to has to be a multiple of 4. Then if we find the square root you would once again get 3*i^(4n+2) which is the exact same number we got previously and could be written as -3. I've been trying to find methods to make inverses like these cancel out the function they are inverses of. I struggle the most with trigonometric functions as it's not really possible to tell from the position on a circle how many times it has revolved around the circle. I wouldn't say that just because two numbers compute to look like the same thing that would mean they automatically are the same thing.

@stanimir5F 6 месяцев назад

I have a question regarding reducing a fraction when it is exponent. Obviously here I am limited by the youtube comments not making a powers so I will have to use ^ as an exponent notation but I am sure you will get the idea Take a look at this example: (-1)^(2/6) If we take the operations one by one -1 to the power of 2 equals 1 and then 6th root should give us ±1 (not counting the complex results) But if we reduce the fraction to 1/3 then cube root of -1 is only -1. So we get 2 different answers based on whether or not we do an action that in other context is considered very standard and legal. So what is the correct approach in this case?

@surajjh2746 6 месяцев назад

First of all when doing questions like these, its best to reduce the fraction hence giving you (-1)^(1/3) = -1. Also evaluating (-1)^(2/6) still only gives you -1 (because of the approach of reducing the exponent). You can see this in a graphing calculator such as desmos and writing the equation y = (x)^(2/6) and seeing the point of intersection at x = -1.

@stanimir5F 6 месяцев назад

@@surajjh2746 the -1 solution is unquestionable. But if we can say that the numerator of the fraction is the powering factor while the denominator is the rooting factor then 2/6 could be interpret as (-1)^2 = 1 and then 6th root of 1 which is ±1. I am not saying this is the correct approach but rather asking if it is :)

@vladislavanikin3398 6 месяцев назад

There are two approaches. If you are working only with real numbers, then a^b for rational (or real) b is defined only for a>0. So that means that (-1)^(2/6) or equivalently (-1)^(1/3) is just not defined. This is, however, not the same as saying that ³√-1 is not defined, the latter is perfectly defined and is equal to -1. So in the end a^(m/n) is equal to m-th power of n-th root of a (or the other way) only for positive (not even zero) bases. The second approach is using complex numbers, but there the problem is with using most of the properties of exponents you are familiar with. And in complex numbers whenever you have a^(mn) you don't always have it being equal to (a^m)^n or (a^n)^m, so the order matters. What that means is that (-1)^(2/6)=(⁶√-1)²≠⁶√(-1)². This ensures that (-1)^(2/6)=(-1)^(1/3) and in both cases you have only three complex values, not six, and no extraneous solution of positive 1.

@stanimir5F 6 месяцев назад

@@vladislavanikin3398 I wasn't aware that (⁶√-1)²≠⁶√(-1)² . Does it mean that we always have to do rooting before the powering when it comes to negative base?

@vladislavanikin3398 6 месяцев назад

@@stanimir5F When it comes to evaluating such expressions with complex numbers? It depends. If the numerator and denominator are coprime, then it doesn't matter, if they aren't, then you either have to reduce or, yes, take the root first. But it's basically only for rationals, in general you just can't use properties of exponents in complex numbers. The most general way of evaluating such an expression is with de Moivre's formula (it's general form).