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When we have an eqaution in the form of $A^2=B^2$ the way that I've been taught would be to square both sides of the equation $\sqrt{A^2} = \sqrt{B^2}$ and the result would be that $A = B$ and $A = -B$ , but why is this the case? $A = B$ seems obvious enough since it "cancels each other out", but we are not allowed to have $-B$ as a result of a $\sqrt x$ sign isn't that right? Because it is supposed to be reserved for only positive numbers right? Or are we allowed to do it because $(-B)^2$ results in $B^2$ and therefore is allowed? I am asking this because we have been taught to do it, but not the reasoning behind it and I would probably make careless mistakes because of it.

emacs drives me nuts
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BroodjeRijst
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  • Yes that the result of $\sqrt\cdot$ is non-negative, but what if $B$ itself is already negative? Then $\sqrt{B^2} = -B$. – peterwhy Aug 10 '22 at 00:03
  • So do you mean that $A^2 = (-B)^2$ and therefore would be $A = -B$ after you square it? But how would that be possible when $(-B)^2 = B$ why would you get the negative? Sorry, I think I need some more clarification :( It does not quite enough sense for me that when you square the equation you also get the negative. I have seeb the other proof using the quadratic equation and that one makes a lot of sense, but my teachers have been telling me to square it and write down $A = B$ and $A = -B$ – BroodjeRijst Aug 10 '22 at 00:16
  • You're taking square root from $A^2$ and $B^2$. Claim that $A^2$ and $B^2$ are positive is not the same as claim that $A$ and $B$ are positive, so $A$ and $B$ can have any sign. If they have the same sign $A^2=B^2$ is equivalent to $A=B$, if they have opposite signs $A^2=B^2$ is equivalent to $A=-B$. In any case for real numbers $\sqrt{A^2}=|A|$, which is not always equal to $A$. Then $A^2=B^2 \Leftrightarrow \sqrt{A^2}=\sqrt{B^2} \Leftrightarrow |A|=|B| \Leftrightarrow A=B \lor A=-B$. – Ivan Kaznacheyeu Aug 10 '22 at 09:02

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A better way to look at this is as follows. If you know that $A^2=B^2$, then what can you say about $A$ and $B$? You know that one of the following is true, either $A=B$ or $A=-B$. You also do not know anything more about $A$ and $B$ than that.

You don't need to think about things cancelling each other out, or about taking square roots of anything, just think of it as a logical proposition.

If $A^2=B^2$ then the only possible logical explanation is that either $A=B$ or $A=-B$. "If your cat had kittens then your cat must be a female." "If your car stops running either you've run out of petrol or the engine is broken". You don't need to keep on tormenting numbers and equations and minus signs like you are doing.

Just to extend this, what do we know about $A$ and $B$ if $A^3=B^3$?

We know that $A=B$. If $A=-B$ then $A^3=-B^3$.

Suzu Hirose
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  • Okay, so the way I see it $A^2 = B^2$ is $AA = BB$ , so $A = B$ makes sense and $AA = (-B)(-B)$ would be $A^2 = (-B)^2$ , which is essentially the same as $A^2 = B^2$ , so therefore is $A = -B$ ? I can understand this reasoning, but I have been wondering why I have been taught to square root both sides and just say $A = B$ or $A = -B$ , I want to know the reasoning behind this one, because it is also the method that comes the most intuitively. – BroodjeRijst Aug 10 '22 at 00:43
  • @BroodjeRijst you are tormenting yourself for no reason. – Suzu Hirose Aug 10 '22 at 00:48
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    +1 for "Just think about it. Don't try to do it with rules." – Ethan Bolker Aug 10 '22 at 02:16
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I think it might be useful to look into a concrete example: find the possible $B$ that would satisfy

$$9 = B^2$$

One case is that $B$ is non-negative, and that corresponds to your "cancel each other out" case:

$$\begin{align*} \sqrt 9 &= \sqrt{B^2}\\ 3&=B \end{align*}$$

Then another case is that $B$ is negative. You are right that the result of $\sqrt{\cdot}$ is non-negative, then if you apply $\sqrt{\phantom{B^2}}$ to $B^2$, the result has to be positive. But among $-B$ and $B$, here $-B$ is the positive number, and $\sqrt{B^2} = -B$:

$$\begin{align*} \sqrt 9 &= \sqrt{B^2}\\ 3 &= -B\\ B &= -3 \end{align*}$$

So $3$ and $-3$ are the only two possible $B$s that satisfy $9=B^2$.

Or in your question, either $A=B$ or $A=-B$.

peterwhy
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  • So lets say that $B = -3$ then $B^2 = (-3)^2$ applying the square root will be $\sqrt B^2 = \sqrt (-3)^2$ , but $\sqrt (-3)^2 = \sqrt ((-3)(-3)) = \sqrt (-3) \sqrt (-3)$ which is illegal right? Idk if I'm going off topic and I apologize if I just don't seem to get it and making you repeat yourself. – BroodjeRijst Aug 10 '22 at 01:30
  • @BroodjeRijst I would like to understand your exact equations in your comment, maybe format the \sqrt line correctly with {}? Especially with the ^2 that can appear outside or inside the \sqrt sign. E.g. \sqrt{(-121)^2} is $\sqrt{(-121)^2}$. – peterwhy Aug 10 '22 at 01:37
  • Apologies, $\sqrt{(-3)^2} = \sqrt{((-3)(-3))} = \sqrt{(-3)} \sqrt{(-3)}$ – BroodjeRijst Aug 10 '22 at 01:48
  • I have been thinking some more and I think this would most probably make the most sense $A^2 = B^2$ gives $\sqrt{(A)^2} = \sqrt{(B)^2}$ and $\sqrt{(A)^2} = -\sqrt{(B)^2}$ , for any A and B . I think if I just look at it this way my problems will be solved. Please feel free to correct me if I'm wrong and I would like to thank you for responding to everything so far. – BroodjeRijst Aug 10 '22 at 01:54
  • @BroodjeRijst Thanks. That's illegal, and also the $\sqrt{(-3)}$ in the RHS is undefined in real numbers. So just to reiterate for completeness, the LHS should be calculated as $\sqrt{(-3)^2} = \sqrt 9 = 3$ (which is the negated $-3$). – peterwhy Aug 10 '22 at 01:54
  • I assume the illegal part was sepparating the $((-3)*(-3))$ because of the extra parentheses around them? – BroodjeRijst Aug 10 '22 at 01:57
  • Separating $(-3)^2 \to (-3)(-3)$ inside the radical sign is OK, but separating $\sqrt{(-3)(-3)} \to \sqrt{-3}*\sqrt{-3}$ is illegal and would cause contradictions like $3=-3$. – peterwhy Aug 10 '22 at 02:00
  • @BroodjeRijst I wouldn't say $\sqrt{(A)^2} = -\sqrt{(B)^2}$, because the LHS is non-negative and the RHS is non-positive, so this is usually false other than $A=B=0$. I think what to note here is that for negative $B$, $\sqrt {B^2}$ is still positive and would $=-B$. Then for $B$ whose sign is unknown, either $\sqrt {B^2} = B$ or $\sqrt {B^2}=-B$. – peterwhy Aug 10 '22 at 02:06
  • Ahh I think I get it now! Oh but about the illegal move, the only reason that it can't split the square root apart is because the number is negative right? In the case that it is positive you can split it apart? – BroodjeRijst Aug 10 '22 at 02:20
  • @BroodjeRijst For all non-negative real numbers $x$ and $y$, $\sqrt{xy}=\sqrt x \sqrt y$. – peterwhy Aug 10 '22 at 02:32
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If you want, there's no need for radicals. $A^2 = B^2$ implies $A^2 - B^2 = 0$ and therefore, $(A - B)(A + B) = 0$. So one of the equations $A - B = 0$ or $A + B = 0$ must be true. Thus $A = B$ or $A = -B$.

on1921379
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