How do you solve $(x+2)^2 = 9$ ?

Question

7458 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-13T16:44:00

$x=1$ or $x=-5$

$(x+2)^2=9$ can be written as $(x+2)^2-9=0$ or

$(x+2)^2-3^2=0$

Now using the identity $a^2-b^2=(a-b)(a+b)$ , we have

$(x+2-3)(x+2+3)=0$ or

$(x-1)(x+5)=0$

Hence either $(x-1)=0$ i.e.

$x=1$

or $(x+5)=0$ i.e. $x=-5$

How do you solve (x+2)^2 = 9(x+2)^2 = 9?