How do you solve by completing the square: $x^{2} + 7 x - 5 = 0$ $x^2+7x-5=0$ ?

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How do you solve by completing the square: $x^{2} + 7 x - 5 = 0$ $x^2+7x-5=0$ ?

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Answer 1 · 2015-04-02T16:40:56

Write it as:
$x^{2} + 7 x = 5$ $x^2+7x=5$
$x^{2} + 7 x + \frac{49}{4} - \frac{49}{4} = 5$ $x^2+7x+49/4-49/4=5$
${(x + \frac{7}{2})}^{2} = 5 + \frac{49}{4}$ $(x+7/2)^2=5+49/4$
${(x + \frac{7}{2})}^{2} = \frac{69}{4}$ $(x+7/2)^2=69/4$
$x + \frac{7}{2} = \pm \sqrt{\frac{69}{4}}$ $x+7/2=+-sqrt(69/4)$
$x + \frac{7}{2} = \pm \frac{\sqrt{69}}{2}$ $x+7/2=+-sqrt(69)/2$
$x = - \frac{7}{2} \pm \frac{\sqrt{69}}{2}$ $x=-7/2+-sqrt(69)/2$

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How do you solve by completing the square: $x^{2} + 7 x - 5 = 0$ $x^2+7x-5=0$ ?

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