How do you factor and solve $r^{2} + 8 r + 14 = 0$ $r^2+8r+14=0$ ?

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How do you factor and solve $r^{2} + 8 r + 14 = 0$ $r^2+8r+14=0$ ?

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Answer 1 · 2016-07-12T14:10:09

$x = - 4 \pm \sqrt{2}$ $x = - 4 +- sqrt2$

Explanation:

Use the improved quadratic formula in graphic form (Google)
$y = r^{2} + 8 r + 14 = 0$ $y = r^2 + 8r + 14 = 0$
$D = d^{2} = b^{2} - 4 a c = 64 - 56 = 8$ $D = d^2 = b^2 - 4ac = 64 - 56 = 8$ --> $d = \pm 2 \sqrt{2}$ $d = +- 2sqrt2$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{8}{2} \pm \frac{2 \sqrt{2}}{2} = - 4 \pm \sqrt{2}$ $x = -b/(2a) +- d/(2a) = -8/2 +- (2sqrt2)/2 = - 4 +- sqrt2$
Factored form of y:
$y = (x - x 1) (x - x 2) = (x + 4 - \sqrt{2}) (x + 4 + \sqrt{2})$ $y = (x - x1)(x - x2) = (x + 4 - sqrt2)(x + 4 + sqrt2)$

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How do you factor and solve $r^{2} + 8 r + 14 = 0$ $r^2+8r+14=0$ ?

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How do you factor and solve r2+8r+14=0r^2+8r+14=0?

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How do you factor and solve $r^{2} + 8 r + 14 = 0$ $r^2+8r+14=0$ ?