How do you solve $8 x^{2} + 32 x - 24$ $8x^2+32x-24$ by completing the square?

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How do you solve $8 x^{2} + 32 x - 24$ $8x^2+32x-24$ by completing the square?

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Answer 1 · 2017-03-20T04:01:44

$x = - 2 + \sqrt{7}, - 2 - \sqrt{7}$ $x = -2 +sqrt 7, -2 - sqrt 7$

Explanation:

$8 x^{2} + 32 x - 24 = 0$ $8 x^2 + 32 x - 24 =0$

reduce coefficient $x^{2}$ $x^2$ to 1 by dividing with $8$ $8$
$x^{2} + 4 x - 3 = 0$ $x^2 +4x -3 = 0$

consider coefficient of $x$ $x$ and divide by 2 then make a parentesis and square them, then square the number in parentesis and deduct it in the equation.
${(x + 2)}^{2} - {(2)}^{2} - 3 = 0$ $(x + 2)^2 - (2)^2 - 3 = 0$
${(x + 2)}^{2} - 4 - 3 = 0$ $(x + 2)^2 - 4 - 3 = 0$
${(x + 2)}^{2} - 7 = 0$ $(x + 2)^2 - 7 = 0$
${(x + 2)}^{2} = 7$ $(x + 2)^2 = 7$
$(x + 2) = \pm \sqrt{7}$ $(x + 2) = +-sqrt 7$
$x = - 2 \pm \sqrt{7}$ $x = -2 +-sqrt 7$

$x = - 2 + \sqrt{7}, - 2 - \sqrt{7}$ $x = -2 +sqrt 7, -2 - sqrt 7$

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How do you solve $8 x^{2} + 32 x - 24$ $8x^2+32x-24$ by completing the square?

1 Answer

Explanation:

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How do you solve $8 x^{2} + 32 x - 24$ $8x^2+32x-24$ by completing the square?