Why does $ \sqrt{x^2} = |x| $

Question

This equality keeps coming up in my algebra course and I don't understand why this is true. $$ \sqrt{x^2} = |x| $$ shouldn't it just be $$ \sqrt{x^2} = x $$?

Answer 1

Remember that every (positive) number has two numbers which square to it: e.g., $$4=2^2=(-2)^2.$$ In order to be a function, the square root operator has to pick one (a function can't be multivalued).

We make the convention that $\sqrt{a}$ is always the nonnegative square root of $a$, so e.g. $\sqrt{4}=2$ and not $-2$. But this means that we need not have $\sqrt{x^2}=x$! Namely, $$\sqrt{(-2)^2}=\sqrt{4}=2\not=-2.$$ Instead, what is true is that $\sqrt{x^2}$ is the "nonnegative version" of $x$ - that is, $\vert x\vert$.

Answer 2

Here's the basic principle of a function: for every input, you get only one output.

Consider $(-9)^2$. This is $81$. If $\sqrt{x^2} = x$, then $\sqrt{81} = \sqrt{(-9)^2} = -9.$

Consider $(9)^2$. This is also $81$. If $\sqrt{x^2} = x$, then $\sqrt{81} = \sqrt{9^2} = 9$.

When I write $\sqrt{81}$, it should unambiguously give me one number, since $\sqrt{\cdot}$ is a function. But I have, from above, that $\sqrt{81} = 9$ and $\sqrt{81} = -9$. This doesn't work if $\sqrt{\cdot}$ is a function.

Thus, we choose $\sqrt{x^2}$ to be the positive (technically, nonnegative) value $|x|$, by convention, so $\sqrt{81} = \sqrt{9^2}$ which also is $\sqrt{(-9)^2}$, but we let $\sqrt{81} = |-9| = |9| = 9$.

Answer 3

No. $\sqrt a$ conventionally denotes the non-negative square root of a non-negative real number, and $x^2\ge 0$ even if $x <0$. Remember a positive real number has two (opposite) square roots.

Answer 4

The square root $\sqrt{y}$ of a non-negative real number $y$ is, by definition, the unique non-negative real number $x$ such that $x^2=y$. So what does this imply for $\sqrt{x^2}$?

If $x$ is not negative, then obviously the non-negative number whose square is the square of $x$ is $x$ itself.

If $x$ is negative, then $-x>0$, and since $(-x)^2=x^2$, we get $\sqrt{x^2}=-x$.

So to summarize, we have $$\sqrt{x^2} = \begin{cases} x & x\ge 0\\ -x & x<0 \end{cases}$$ But that is exactly the definition of $|x|$.

Answer 5

Consider the three cases where $x<0$, $x=0$ and $x>0$ separately. In each case, you'll find that $\sqrt{x^2}=\vert x \vert$.

Answer 6

This follows from the definition of absolute value: If $0\leq a$, then $\sqrt{a}$ is defined as the non-negative unique solution of the equation $x^2=a$. Now, since the solutions of the equation $x^2=a^2$ are $x=|a|$ and $a=-|a|$, it follows by the definition that $\sqrt{a^2}=|a|$.