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

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Answer 1 · 2016-07-13T16:44:00

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

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

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

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

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

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

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

$x = 1$ $x=1$

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

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