How do you solve $x^{2} = - 5 x - 5$ $x^2 = -5x - 5$ ?

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How do you solve $x^{2} = - 5 x - 5$ $x^2 = -5x - 5$ ?

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Answer 1 · 2016-06-10T03:07:08

$x = \frac{- 5 \pm \sqrt{5}}{2}$ $x = (-5 +- sqrt5)/2$

Explanation:

First, bring the equation to standard form:
$y = x^{2} + 5 x + 5 = 0$ $y = x^2 + 5x + 5 = 0$
Use the improved quadratic formula (Google Search)
$D = d^{2} = b^{2} - 4 a c = 25 - 20 = 5$ $D = d^2 = b^2 - 4ac = 25 - 20 = 5$ --> $d = \pm \sqrt{5}$ $d = +- sqrt5$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{5}{2} \pm \frac{\sqrt{5}}{2} = \frac{- 5 \pm \sqrt{5}}{2}$ $x = -b/(2a) +- d/(2a) = -5/2 +- sqrt5/2 = (-5 +- sqrt5)/2$

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How do you solve $x^{2} = - 5 x - 5$ $x^2 = -5x - 5$ ?

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