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

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?

• It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $\sqrt{\ }$, but after solving the resulting inequality, you need to check for false solutions. – user647486 Apr 27 '19 at 14:41
• Actually the solution should be $25 \geq 5 - x \geq 0 \implies -20 \le x \le 5$ – leonbloy Apr 28 '19 at 3:19

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.

• You don't really "know that" $5-x \ge 0$, rather you "must have" $5-x \ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 \ge |5-x|$ becomes $25 \ge 5-x$. – steven gregory Apr 27 '19 at 14:42
• @stevengregory - I am curious what this great difference of meaning between "know that" and "must have" is. To me in this context, they are synonymous. – Paul Sinclair Apr 27 '19 at 16:42

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

Try with $$a=-1$$.

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

• Indeed: Writing $\sqrt{a}$ implies that $a \geq 0$ if $a$ is a real number – Barranka Apr 27 '19 at 20:41
• Yes, I know. But see the original context. This was applied to the inequality $5\geqslant\sqrt{5-x}$. The OP thought the one could conclude from it that $5^2\geqslant\lvert5-x\rvert$, whereas the correct conclusion is that $5-x\geqslant0$ and that $5^2\geqslant5-x$. – José Carlos Santos Apr 27 '19 at 20:45
• But in that case $5-x = |5-x|$, so OP also isn't 'incorrect', per se, just missing an additional detail. – Hayden Apr 27 '19 at 22:03

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

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

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 $$(\sqrt{a})^2 = (\sqrt{-1})^2 = i^2 = -1$$