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

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Answer 1 · 2016-03-31T17:29:56

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

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

$x^{2} + 2 x = 8$ $x^2 + 2x = 8$

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

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

$x^{2} + 2 \cdot x \cdot 1 + 1 = 9$ $x^2 + 2* x * 1 + 1 = 9$

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

${(x + 1)}^{2} = 9$ $(x+1)^2 = 9$

$x + 1 = \sqrt{9}$ $x + 1 = sqrt9$ or $x + 1 = - \sqrt{9}$ $x + 1 = -sqrt9$

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

How do you solve using the completing the square method x2+2x−8=0x^2+2x-8=0?