How do you solve $x^{2} + x - 6 = 0$ $x^2+x-6=0$ by completing the square?

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How do you solve $x^{2} + x - 6 = 0$ $x^2+x-6=0$ by completing the square?

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Answer 1 · 2017-07-02T02:35:56

$x = 2$ $x=2$ or $- 3$ $-3$

Explanation:

Moving the $- 6$ $-6$ to the right side:
$x^{2} + x = 6$ $x^2+x=6$

Completing the square by adding ${(\frac{b}{2})}^{2}$ $(b/2)^2$ to both sides, where $b$ $b$ is the second coefficient (in this case, $b = 1$ $b=1$ ):
$x^{2} + x + {(\frac{1}{2})}^{2} = 6 + {(\frac{1}{2})}^{2}$ $x^2+x+(1/2)^2=6+(1/2)^2$
$x^{2} + x + \frac{1}{4} = 6 + \frac{1}{4}$ $x^2+x+1/4=6+1/4$
${(x + \frac{1}{2})}^{2} = \frac{25}{4}$ $(x+1/2)^2=25/4$

Simplifying:
$x + \frac{1}{2} = \pm \frac{5}{2}$ $x+1/2=\pm5/2$
$x = - \frac{1}{2} \pm \frac{5}{2}$ $x=-1/2\pm5/2$
$x = \frac{4}{2}$ $x=4/2$ or $- \frac{6}{2}$ $-6/2$
$x = 2$ $x=2$ or $- 3$ $-3$

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How do you solve $x^{2} + x - 6 = 0$ $x^2+x-6=0$ by completing the square?

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How do you solve $x^{2} + x - 6 = 0$ $x^2+x-6=0$ by completing the square?