Question #dd07c

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Question #dd07c

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Answer 1 · 2016-04-22T16:25:52

$x = 2, - 1$ $x=2, -1$

Explanation:

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

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

$x^{2} - x = 2$ $x^2-x=2$

Add the square of -1/2 (the coefficient of the $- x$ $-x$ divided by 2) to both sides. Remember, the standard form is $x^{2} + 2 a x + a^{2}$ $x^2 + 2ax + a^2$
$x^{2} - x + \frac{1}{4} = 2 + \frac{1}{4}$ $x^2-x+1/4=2+1/4$

Factor the left side of the equation
$(x - \frac{1}{2}) (x - \frac{1}{2}) = 2 \frac{1}{4}$ $(x-1/2)(x-1/2)=2 1/4$
${(x - \frac{1}{2})}^{2} = 2 \frac{1}{4}$ $(x-1/2)^2 =2 1/4$

$\sqrt{{(x - \frac{1}{2})}^{2}} = \sqrt{2 \frac{1}{4}}$ $sqrt((x-1/2)^2) =sqrt(2 1/4)$

$(x - \frac{1}{2}) = \pm 1 \frac{1}{2}$ $(x-1/2)=+-1 1/2$

$x = \frac{1}{2} \pm 1 \frac{1}{2} = 2, - 1$ $x=1/2+-1 1/2=2,-1$

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Question #dd07c

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