If $25 = a^2$ then the value of a will be

Question

My question is if $25 = a^2$ then the value of $a$ will be?

We always do it as if $$ 25 = a^2 \implies \sqrt{25}= \sqrt{a^2} \implies \pm 5 = a. $$ However why don't we do it as $$ 25 = a^2 \implies \sqrt{25}=\sqrt{a^2} \implies \pm 5 = \pm a $$ Because $(+a )\times( +a) = a^2$ and also $(-a) \times( -a )= a^2$?

Answer 1

Hint: write your equation as $$(5-a)(5+a)=0$$

Answer 2

To offer a different perspective to the comments, the plus-minus ($\pm$) before the $a$ is actually redundant. Observe that if you do all the plus-minus combinations you get $$ \begin{array}{llll} (1) & 5 = a & (2) & -5 = a \\ (3) & 5 = -a & (4) & -5 = -a \\ \end{array} $$ But $(1)$ and $(4)$ are the same (check), and similarly $(2)$ and $(3)$ are the same (check). You get too many of the same answers with the second plus-minus.

Answer 3

According to the definition of square root, the correct way to solve should be

$$a^2=25\iff \sqrt{a^2}=\sqrt{25}\iff |a|=5 $$

then $a=\pm 5$.

Answer 4

$a=5$ and $a=-5$ are the solution of this equation. If I take $y= 5$ And I take square on both sides .then,we get $y^2=25$ After solving the quadratic equation. I get ,$y=\pm 5$.

But I doesn't take $y=-5$.

This type of solution is called a extraneous solution