How do you solve $x^{2} + 4 x - 9 = 0$ $x^2 + 4x - 9 = 0$ by completing the square?

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How do you solve $x^{2} + 4 x - 9 = 0$ $x^2 + 4x - 9 = 0$ by completing the square?

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Answer 1 · 2017-01-11T21:36:09

$x = - 2 \pm \sqrt{13}$ $x = -2+-sqrt(13)$

Explanation:

The difference of squares identity can be written:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

Hence we find:

$0 = x^{2} + 4 x - 9$ $0 = x^2+4x-9$

$0 = x^{2} + 4 x + 4 - 13$ $color(white)(0) = x^2+4x+4-13$

$0 = {(x + 2)}^{2} - {(\sqrt{13})}^{2}$ $color(white)(0) = (x+2)^2-(sqrt(13))^2$

$0 = ((x + 2) - \sqrt{13}) ((x + 2) + \sqrt{13})$ $color(white)(0) = ((x+2)-sqrt(13))((x+2)+sqrt(13))$

$0 = (x + 2 - \sqrt{13}) (x + 2 + \sqrt{13})$ $color(white)(0) = (x+2-sqrt(13))(x+2+sqrt(13))$

Hence:

$x = - 2 \pm \sqrt{13}$ $x = -2+-sqrt(13)$

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How do you solve $x^{2} + 4 x - 9 = 0$ $x^2 + 4x - 9 = 0$ by completing the square?

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Explanation:

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Humanities

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How do you solve x^2 + 4x - 9 = 0x2+4x−9=0x^2 + 4x - 9 = 0 by completing the square?

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Explanation:

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How do you solve $x^{2} + 4 x - 9 = 0$ $x^2 + 4x - 9 = 0$ by completing the square?