How do you factor and solve $9 x^{2} + 7 x - 56 = 0$ $9x^2 + 7x - 56 = 0$ ?

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How do you factor and solve $9 x^{2} + 7 x - 56 = 0$ $9x^2 + 7x - 56 = 0$ ?

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Answer 1 · 2016-07-04T03:30:21

(-7 +- sqrt2065)/18

Explanation:

$f (x) = 9 x^{2} + 7 x - 56 = 0$ $f(x) = 9x^2 + 7x - 56 = 0$
$D = d^{2} = b^{2} - 4 a c = 49 + 2016 = 2065$ $D = d^2 = b^2 - 4ac = 49 + 2016 = 2065$
Since D is not a perfect square, this function can't be factored.
Solve f(x) = 0 by the improved quadratic formula.
$D = 2065$ $D = 2065$ --> $d = \pm \sqrt{2065}$ $d = +- sqrt(2065)$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{7}{18} \pm \frac{\sqrt{2065}}{18} = \frac{- 7 \pm \sqrt{2065}}{18}$ $x = -b/(2a) +- d/(2a) = - 7/18 +- sqrt(2065)/18 = (-7 +- sqrt(2065))/18$

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How do you factor and solve $9 x^{2} + 7 x - 56 = 0$ $9x^2 + 7x - 56 = 0$ ?

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How do you factor and solve $9 x^{2} + 7 x - 56 = 0$ $9x^2 + 7x - 56 = 0$ ?