Does the $\sqrtx^2$ equal $x$ or $|x|$? In various other words is the an agreement that us take by default the hopeful square root of $x$ or we need to explicitly specify it v absolute worth as there might be two roots to $x^2$ ($+x$ or $-x$)?




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Assuming you room talking about real $x$, the square root is defined to be the primary square root i.e. The positive one. Therefore $\sqrtx^2=|x|$, as, for example $\sqrt4=2$ and also not $-2$, while $(-2)^2=4$


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$\sqrt x^2\geq 0$ for every $x$ and likewise $|x|=x$ if $x\geq 0$ and also $|x|=-x$ if $x\leq 0$ and thus $\sqrt x^2 =x$ if $x\geq 0$

and $\sqrt x^2 =-x $if $x\leq 0$ or $\sqrt x^2 =|x|$


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The prize $\sqrtx$ always denotes the positive number $y$ satisfying the equation $y^2 = x$. The thinking behind this, as far as ns know, is to make $\sqrtx$ a function. Every optimistic real number has actually two square roots (e.g. Both $3$ and also $-3$ space square root of $9$), however if $\sqrtx$ denoted both of this numbers, then it would be much more annoying to transaction with. Because that example, in an inequality, you would not be able to multiply both political parties by $\sqrtx$, due to the fact that you would have no guarantee the $\sqrtx > 0$. By phone call $\sqrtx$ the confident root, it to represent one number, no two, and this enables you to only deal with the an unfavorable root when you desire to.

To be regular with $\sqrtx > 0$, we have to say $\sqrtx^2 = |x|$.

To reinforce this, consider $\sqrt(-2)^2$. If we simply said the $\sqrtx^2 = x$, then we would have $\sqrt(-2)^2 = -2$. While it"s true that $-2$ is a square source of $4$, because $(-2)^2 = 4$, the is not the positive square root, therefore this prize is no what us want. The optimistic square source is $2$, and since $\sqrtx$ denotes the optimistic root, $\sqrt(-2)^2 = |-2| = 2.$