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I was trying a problem and was getting the wrong answer and when I saw the solution on the internet I found this statement written in square brackets $\sqrt{x^2}$[note square is on $x$] is $|x|$. Till now I have learned that by laws of exponents we can multiply the powers and obtain $\sqrt{x}^2$ as $x$. But this seems confusing to me. Please clear this doubt.

And does this apply to $\left({\sqrt x}\right)^2$[note square is on radical] i.e. is $\left({\sqrt x}\right)^2 = |x|$?

Sahiba Arora
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Shreyansh Kuntal
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    What is $(-1)^2$ ? What is $\sqrt1$ ? Then what is $\sqrt{(-1)^2}$ ? –  Jul 11 '20 at 12:45
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    "Till now I have learnt that by laws of exponents" What exactly do those laws say? Because I think you missed a small, but very important detail. – Arthur Jul 11 '20 at 12:45
  • @Arthur can you tell what is that. – Shreyansh Kuntal Jul 11 '20 at 12:47
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    @ShreyanshKuntal: if you don't tell how you interpret the "laws of exponent", we cannot tell you why you are wrong. –  Jul 11 '20 at 12:48
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    If non-integer exponents are involved, the base must be non-negative , otherwise the law $(a^b)^c=a^{bc}$ does not hold in general. In fact, within the reals $a^b$ can be undefined. The reason that $\sqrt{x^2}$ is not just $x$ is that the convention of a square-root is that the non-negative possibility is meant, and for negative $x$, we get therefore $-x$. If we start with $\sqrt{x}$ , we must (at least in the reals) assume that $x\ge 0$ and then $(\sqrt{x})^2=x$ follows from the exponential law. – Peter Jul 11 '20 at 12:56
  • @Peter That's really an answer, not a comment. In fact, my above comment is borderline, in my opinion. – Arthur Jul 11 '20 at 12:59
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    @Arthur There are already $3$ answers. Usually, I do not add another answer in this case, since usually everything has been said. But I think, it is a useful summary. – Peter Jul 11 '20 at 13:04
  • @Peter then can we say [(x)^(1/n)]^n is not equal to x. – Shreyansh Kuntal Jul 11 '20 at 13:30
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    For $x\ge 0$ , it surely is equal $x$, but for negative $x$, we can have problems. However, if $n$ is an odd positive integer, it works, since the $n$ th root is negative and the equation therefore valid.. The reason is that $f(x)=x^{\frac{1}{n}}$ is bijective in this case, so also negative values work. – Peter Jul 11 '20 at 13:37

4 Answers4

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The square root of $x$ is a non-negative number $y$ such that $y^2=x$. Thus the square root of $x^2$ is a non-negative number $y$ such that $y^2=x^2$. To make sure the square root of $x^2$ is non-negative then you take $y=\lvert x\rvert$.

In the second part, writing $(\sqrt{x})^2$ means that $x \geq 0$, so $\lvert x\rvert = x$, and you can safely say $(\sqrt{x})^2 = x$.

Gibbs
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We know that, for any $x\in \Bbb R^+$ there exist two $y\in\Bbb R : y^2=x$, namely $y=y_+$ and $y=y_-=-y_+$.

Therefore, the inverse of the function $ f:\Bbb R\mapsto \Bbb R :f(x)=x^2$ will be one-to-many (i.e. $f^{-1}(4)=2$ and $f^{-1}(4)=-2$ simultaneously)

This breaks the definition of a function, which must be one-to-one or many-to-one.

So we define $f^{-1}(x)=\sqrt x=|y|=y_+$ so that the function is one-to-one and therefore valid.

Rhys Hughes
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Partial answer:

$$\left(\sqrt x\right)^2=|x|$$ obviously holds because the expression is only defined for $x\ge0$. For the same reason,

$$\left(\sqrt x\right)^2=x$$ holds.

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By definition $\sqrt{x^2}$ is (the unique) non-negative number $a$
such that $a^2 = x^2$. I guess you will agree with this.

Now... if we take $|x|$ we see that this number $|x|$ satisfies both
conditions for $a$ i.e. it is non-negative and its square is $x^2$. So $|x| = a$

peter.petrov
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