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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 $(-a) \times( -a )= a^2$?

Awesome boy
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  • Both $5$ and $-5$ solve the given equation. By definition, $\sqrt x$ is the positive square root of $x$. – lulu May 13 '18 at 10:29
  • The solutions of $25=a^2$ are indeed $a=5$ and $a=-5$. Note that for all real $x$, we have $\sqrt{x^2}=|x|$, hence applying the square root function to both sides of $25=a^2$ we arrive at $5=|a|$, in other words $a=5$ or $a=-5$. – Hagen von Eitzen May 13 '18 at 10:29
  • See related post. – farruhota May 13 '18 at 10:36
  • Why did you add modulus to x – Awesome boy May 13 '18 at 10:50
  • @Awesomeboy Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 31 '18 at 21:00

4 Answers4

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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.

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

Dr. Sonnhard Graubner
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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 $a=\pm 5$.

user
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$a=5$ 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