How do you solve by completing the square for $2 x^{2} + 7 x - 15 = 0$ $2x^2+7x - 15 = 0$ ?

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How do you solve by completing the square for $2 x^{2} + 7 x - 15 = 0$ $2x^2+7x - 15 = 0$ ?

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Answer 1 · 2015-03-30T17:47:23

Group the two variable terms together, then factor out the number in front of $x^{2}$ $x^2$

$(2 x^{2} + 7 x) - 15 = 0$ $(2x^2+7x)-15=0$

$2 (x^{2} + \frac{7}{2} x + leave space) - 15 = 0$ $2(x^2+7/2 x+ color(white)"leave space") -15=0$

Take $\frac{1}{2}$ $1/2$ of the coefficient of $x$ $x$ , square that and add and subtract inside the parentheses:

$\frac{1}{2} of \frac{7}{2} = \frac{7}{4}$ $1/2 " of " 7/2 = 7/4$ Squaring gives us $\frac{49}{16}$ $49/16$ , so we write:

$2 (x^{2} + \frac{7}{2} x + \frac{49}{16} - \frac{49}{16}) - 15 = 0$ $2(x^2+7/2 x+ 49/16 - 49/16) -15=0$ Keep the positive $\frac{49}{16}$ $49/16$ inside the parentheses (we need it for the perfect square)

Write:
$2 (x^{2} + \frac{7}{2} x + \frac{49}{16}) - 2 (\frac{49}{16}) - 15 = 0$ $2(x^2+7/2 x+ 49/16) - 2(49/16) -15=0$

Factor the square and simplify the rest:

$2 {(x + \frac{7}{4})}^{2} - \frac{49}{8} - \frac{120}{8} = 0$ $2(x + 7/4)^2 - 49/8 - 120/8=0$

$2 {(x + \frac{7}{4})}^{2} - \frac{169}{8} = 0$ $2(x + 7/4)^2 - 169/8=0$

$2 {(x + \frac{7}{4})}^{2} = \frac{169}{8}$ $2(x + 7/4)^2 = 169/8$

${(x + \frac{7}{4})}^{2} = \frac{169}{16}$ $(x + 7/4)^2 = 169/16$

$x + \frac{7}{4} = \pm \sqrt{\frac{169}{16}}$ $x + 7/4 = +- sqrt(169/16)$

$x + \frac{7}{4} = \pm \frac{13}{4}$ $x + 7/4 = +- 13/4$

$x = - \frac{7}{4} \pm \frac{13}{4}$ $x = -7/4 +- 13/4$

$- \frac{7}{4} + \frac{13}{4} = \frac{6}{4} = \frac{3}{2}$ $-7/4 + 13/4 = 6/4=3/2$ and $- \frac{7}{4} - \frac{13}{4} = - \frac{20}{4} = - 5$ $-7/4 - 13/4 = -20/4 = -5$

The solutions are $\frac{3}{2}$ $3/2$ , $5$ $5$

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How do you solve by completing the square for $2 x^{2} + 7 x - 15 = 0$ $2x^2+7x - 15 = 0$ ?

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