How do you solve using the completing the square method $x^2+2x-8=0$ ?

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Answer 1 · 2016-03-31T17:29:56

The solutions are:
$color(green)(x = 2$ ,
$color(green)(x = -4$

$x^2 + 2x- 8 =0$

$x^2 + 2x = 8$

To write the Left Hand Side as a Perfect Square, we add 1 to both sides:

$x^2 + 2x + 1 = 8 + 1$

$x^2 + 2* x * 1 + 1 = 9$

Using the Identity $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get

$(x+1)^2 = 9$

$x + 1 = sqrt9$ or $x + 1 = -sqrt9$

$color(green)(x = 3 -1 = 2$ or $color(green)(x = -3 - 1 = -4$

How do you solve using the completing the square method x^2+2x-8=0x^2+2x-8=0?