How do you solve $9 x^{2} - 7 x = 12$ $9x^2 - 7x = 12$ using completing the square?

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How do you solve $9 x^{2} - 7 x = 12$ $9x^2 - 7x = 12$ using completing the square?

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Answer 1 · 2015-06-19T00:27:03

I found:
$x_{1} = \frac{1}{18} (7 + \sqrt{481})$ $x_1=1/18(7+sqrt(481))$
$x_{2} = \frac{1}{18} (7 - \sqrt{481})$ $x_2=1/18(7-sqrt(481))$

Explanation:

Let us do some manipulations:

$9 x^{2} - 7 x = 12$ $9x^2-7x=12$ add and subtract $\frac{49}{36}$ $49/36$
$9 x^{2} - 7 x + \frac{49}{36} - \frac{49}{36} = 12$ $9x^2-7x+49/36-49/36=12$
$9 x^{2} - 7 x + \frac{49}{36} = 12 + \frac{49}{36}$ $9x^2-7x+49/36=12+49/36$
${(3 x - \frac{7}{6})}^{2} = \frac{481}{36}$ $(3x-7/6)^2=481/36$ root square both sides:
$3 x - \frac{7}{6} = \pm \sqrt{\frac{481}{36}} =$ $3x-7/6=+-sqrt(481/36)=$
$x = \frac{1}{3} (\frac{7}{6} \pm \sqrt{\frac{481}{36}})$ $x=1/3(7/6+-sqrt(481/36))$
$x_{1} = \frac{1}{18} (7 + \sqrt{481})$ $x_1=1/18(7+sqrt(481))$
$x_{2} = \frac{1}{18} (7 - \sqrt{481})$ $x_2=1/18(7-sqrt(481))$

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How do you solve $9 x^{2} - 7 x = 12$ $9x^2 - 7x = 12$ using completing the square?

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