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Find the solution to the following equation by transforming it into a perfect square trinomial.* x2 + 2x = 8

Sagot :

x2+2x-8 =0
x2 + 4x - 2x - 8 = 0
x(x+4) - 2(x+4) = 0
(x-2)(x+4) = 0
let x- 2 = 0
x = 2
let x+4 = 0
x= -4
Ryan2
[tex] x^{2} +2x=8 \ add \ 1 \ to \ both \ sides\\ \\ x^{2} +x+8+1 \ in \ the \ left \ side \ you \ find \ a perfect \ square \ trinomial\\ \\ (x+1)^2=9\\ \\ So:\\ \\ x+1=\pm \sqrt{9}\\ \\ x+1=\pm 3\\ \\ x+1=3 \rightarrow x=2\\ \\ x+1=-3 \rightarrow \x=-4[/tex]
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