How do you solve $2 x^{2} - 2 x + 7 = 5$ $2x^2-2x+7=5$ by completing the square?

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How do you solve $2 x^{2} - 2 x + 7 = 5$ $2x^2-2x+7=5$ by completing the square?

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Answer 1 · 2017-07-19T04:18:52

$x = \frac{1 \pm i \sqrt{3}}{2}$ $x = (1 +- isqrt3)/2$

Explanation:

$2 x^{2} - 2 x = - 2$ $2x^2 - 2x = - 2$
$x^{2} - x = - 1$ $x^2 - x = - 1$
$x^{2} - x + \frac{1}{4} = - 1 + \frac{1}{4}$ $x^2 - x + 1/4 = - 1 + 1/4$
${(x - \frac{1}{2})}^{2} = - \frac{3}{4} = \frac{3 i^{2}}{4}$ $(x - 1/2)^2 = - 3/4 = (3i^2)/4$
There are 2 complex roots:
$x - \frac{1}{2} = \pm \frac{i \sqrt{3}}{2}$ $x - 1/2 = +- (isqrt3)/2$
$x = \frac{1}{2} \pm \frac{i \sqrt{3}}{2} = \frac{1 \pm i \sqrt{3}}{2}$ $x = 1/2 +- (isqrt3)/2 = (1 +- isqrt3)/2$

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How do you solve $2 x^{2} - 2 x + 7 = 5$ $2x^2-2x+7=5$ by completing the square?

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