How do you solve $x^{2} + 3 = 9 x$ $x^2 + 3 = 9x$ using the quadratic formula?

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How do you solve $x^{2} + 3 = 9 x$ $x^2 + 3 = 9x$ using the quadratic formula?

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Answer 1 · 2016-05-11T03:46:04

$x = \frac{9 \pm \sqrt{69}}{2}$ $x = (9 +- sqrt69)/2$

Explanation:

$y = x^{2} - 9 x + 3 = 0 .$ $y = x^2 - 9x + 3 = 0.$
Use the improved quadratic formula (Google, Yahoo Search).
$D = d^{2} = b^{2} - 4 a c = 81 - 12 = 69$ $D = d^2 = b^2 - 4ac = 81 - 12 = 69$ --> $d = \pm \sqrt{69}$ $d = +- sqrt69$ .
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{9}{2} \pm \frac{\sqrt{69}}{2} = \frac{9 \pm \sqrt{69}}{2}$ $x = -b/(2a) +- d/(2a) = 9/2 +- sqrt69/2 = (9 +- sqrt69)/2$

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How do you solve $x^{2} + 3 = 9 x$ $x^2 + 3 = 9x$ using the quadratic formula?

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How do you solve $x^{2} + 3 = 9 x$ $x^2 + 3 = 9x$ using the quadratic formula?