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I have a question here... Usually, for
$x^2 = 4$
$x=\sqrt{4}$
$x=±2$

But if the question is like this :
$y^2 = (x+2)(x+2)$
$y^2 = (x+2)^2$

If I want to find $y$ in term of $x$,I will put square root on both sides.

y = ±(x+2)

So I'm wondering about whether I should put the sign ± in front of $(x+2)$.This because I think the answer should be $(x+2)$ only because the $(x+2)^2$ is the product of $(x+2)(x+2)$ as shown in the equation and not $(-（x+2)) (-(x+2))$. Anyone can tell me ? Thank you :D

Steven Alexis Gregory
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user30356
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  • Welcome at MSE, please use Mathjax/Latex Syntax. $y=\pm(x+2)$ is correct. – MrYouMath Jun 19 '16 at 12:53
  • If you have been taught that $\sqrt 4 = \pm 2$, unlearn it. $\sqrt 4$ is defined to be the positive square root of $4$, that is $\sqrt 4 = 2$. So the correct response to $x^2 = 4$ would be $x = \pm \sqrt 4$. – Steven Alexis Gregory Jun 19 '16 at 13:19
  • If you had $y^2 = (-7)(-7) = (-7)^2$, then would you want to say $y = -7$? It is a mistake to treat the expression $x+2$ as if it were a positive number. – Steven Alexis Gregory Jun 19 '16 at 13:26

1 Answers1

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Actually, the solution is $\pm (x+2)$ (as MrYouMath pointed out in the comments). This is because if you multiply $-(x+2)$ by itself, you get $(-(x+2))(-(x+2))$, which is equivalent to $(-1)(-1)(x+2)(x+2)$ by the commutative laws of multiplication. Finally, the first product is $-1 \times -1=1$, so indeed you would get $(-(x+2))^2=(x+2)^2=y^2$, meaning both are a solution to your equation.

exfret
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