How do you solve the equation $2 x^{2} + 10 x = - 17$ $2x^2+10x=-17$ by completing the square?

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How do you solve the equation $2 x^{2} + 10 x = - 17$ $2x^2+10x=-17$ by completing the square?

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Answer 1 · 2018-03-03T10:23:13

$x =$ $color(red)(x =$ $- (\frac{1}{2}) (5 - 3 i), (- \frac{1}{2}) (5 + 3 i)$ $color(green)(-(1/2)(5-3i), (-1/2)(5+3i)$

Explanation:

$2 x^{2} + 10 x = - 17$ $2x^2 + 10x = -17$

Divide by 2 on both sides. $x^{2} + 5 x = - \frac{17}{2}$ $x^2 + 5x = -17/2$

Rewrite the (xy term as multiple of 2#

$x^{2} + (2 \cdot x \cdot (\frac{5}{2})) = - \frac{17}{2}$ $x^2 + (2 * x * (5/2)) = -17/2$

Complete the square on L H S by adding ${(\frac{5}{2})}^{2}$ $(5/2)^2$ to both sides.

$x^{2} + (2 \cdot (\frac{5}{2}) x + {(\frac{5}{2})}^{2} = - \frac{17}{2} + {(\frac{5}{2})}^{2} = - \frac{9}{4}$ $x^2 + (2*(5/2) x+ (5/2)^2 = -17 / 2 + (5/2)^2 = -9/4$

${(x + \frac{5}{2})}^{2} = ({\sqrt{- \frac{9}{4}}}^{2} = (i \cdot {(\frac{3}{2})}^{2}$ $(x+5/2)^2 = (sqrt(-9/4)^2 = (i * (3/2)^2$

$x + \frac{5}{2} = \pm (\frac{3}{2}) i$ $x + 5/2 = +- (3/2)i$

$x = - \frac{5}{2} + \frac{3}{2} i, - \frac{5}{2} - \frac{3}{2} i$ $color(red)(x ) = color(blue)(-5/2 + 3/2i, -5/2 -3/2i$ or $- (\frac{1}{2}) (5 - 3 i), (- \frac{1}{2}) (5 + 3 i)$ $color(green)(-(1/2)(5-3i), (-1/2)(5+3i)$

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How do you solve the equation $2 x^{2} + 10 x = - 17$ $2x^2+10x=-17$ by completing the square?

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Explanation:

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How do you solve the equation $2 x^{2} + 10 x = - 17$ $2x^2+10x=-17$ by completing the square?