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

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

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Answer 1 · 2015-06-27T11:35:38

Ifound:
$x_{1} = 2$ $x_1=2$
$x_{2} = - 4$ $x_2=-4$

Explanation:

Take $- 8$ $-8$ to the right:
$x^{2} + 2 x = 8$ $x^2+2x=8$
Add and subtract $1$ $1$ :
$x^{2} + 2 x + 1 - 1 =$ $x^2+2x+1-1=$
$x^{2} + 2 x + 1 = 8 + 1$ $x^2+2x+1=8+1$
${(x + 1)}^{2} = 9$ $(x+1)^2=9$
Root square both sides:
$x + 1 = \pm \sqrt{9}$ $x+1=+-sqrt(9)$
$x + 1 = \pm 3$ $x+1=+-3$
$x = - 1 \pm 3$ $x=-1+-3$
$x_{1} = 2$ $x_1=2$
$x_{2} = - 4$ $x_2=-4$

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

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Explanation:

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

1 Answer

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