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 · 2015-08-01T18:16:38

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

Explanation:

$x^{2} = - 5 x - 5$ $x^2 = -5x-5$

Convert the equation to standard form,

$a x^{2} + b x + c = 0$ $ax^2 + bx +c =0$

$x^{2} + 5 x + 5 = 0$ $x^2 + 5x+5 = 0$

$a = 1$ $a=1$ , $b = 5$ $b=5$ , and $c = 5$ $c=5$

Now use the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 5 \pm \sqrt{5^{2} - 4 \times 1 \times 5}}{2 \times 1} = \frac{- 5 \pm \sqrt{25 - 20}}{2} = \frac{- 5 \pm \sqrt{5}}{2}$ $x = (-b±sqrt(b^2 -4ac))/(2a) = (-5±sqrt(5^2-4×1×5))/(2×1) = (-5±sqrt(25-20))/2 = (-5±sqrt5)/2$

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

or

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

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

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