Why isn't the definition of absolute value applied when squaring a radical containing a variable?

Question

I recently learned about the following definition of absolute value:

$|a| = \sqrt{a^2}$

Then I came across a solution to a problem that had the following step:

$5 \geq \sqrt{5 - x}$

In order to proceed, we had to square both sides:

$5^2 \geq (\sqrt{5 - x})^2$

With the aforementioned definition of absolute value in mind, I wrote:

$25 \geq |5 - x|$

But the actual solution turned out to be:

$25 \geq 5 - x$

I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?

Answer 1

From the fact that you can take $\sqrt {5-x}$ you know that $5-x \ge 0$ so you don't need the absolute value signs.

Answer 2

$$\left(\sqrt a\right)^2\ne\sqrt{a^2}.$$

Try with $a=-1$.

Answer 3

Indeed, $\sqrt{a^2}=\lvert a\rvert$. But $\sqrt a^2=a$ (assuming that $a\geqslant0$), not $\lvert a\rvert$.

Answer 4

To compare $\sqrt{5-x}$ to $\sqrt{a^2}$ you must compare $5-x$ to $a^2$. The problem is that $a^2 \ge 0$ while $5-x$ can be any real number. But if you add the restriction $5-x \ge 0$, then $25 \geq |5 - x|$ becomes

$$\text{$25 \geq 5 - x \ \text{and} \ 5-x \ge 0$}$$

Answer 5

If you look carefully, you'll notice your definition has the square inside the square root (not outside): $$|a| = \sqrt{a^2}$$ However, in your solution you seem to assume that: $$\sqrt{a^2} = (\sqrt{a})^2$$ However, $$\sqrt{a^2} \neq (\sqrt{a})^2$$ For example, as others have suggested, if $a = -1$ we have: $$\sqrt{a^2} = \sqrt{(-1)^2} = \sqrt{1} = 1$$ but $$(\sqrt{a})^2 = (\sqrt{-1})^2 = i^2 = -1$$