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 $\left({\sqrt x}\right)^2 = |x|$?
On
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.
On
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 $|x| = a$
On
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.
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 $(\sqrt{x})^2 = x$.