How do you solve $x^{2} + 2 x - 120 = 0$ $x^2+2x-120=0$ by completing the square?

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How do you solve $x^{2} + 2 x - 120 = 0$ $x^2+2x-120=0$ by completing the square?

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Answer 1 · 2016-10-11T06:17:54

$x = 10$ $x=10$ or $x = - 12$ $x=-12$

Explanation:

The difference of squares identity can be written:

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

Use this with $a = (x + 1)$ $a=(x+1)$ and $b = 11$ $b=11$ as follows:

$0 = x^{2} + 2 x - 120$ $0 = x^2+2x-120$

$0 = x^{2} + 2 x + 1 - 121$ $color(white)(0) = x^2+2x+1-121$

$0 = {(x + 1)}^{2} - 11^{2}$ $color(white)(0) = (x+1)^2-11^2$

$0 = ((x + 1) - 11) ((x + 1) + 11)$ $color(white)(0) = ((x+1)-11)((x+1)+11)$

$0 = (x - 10) (x + 12)$ $color(white)(0) = (x-10)(x+12)$

Hence roots $x = 10$ $x=10$ and $x = - 12$ $x=-12$

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How do you solve $x^{2} + 2 x - 120 = 0$ $x^2+2x-120=0$ by completing the square?

1 Answer

Explanation:

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How do you solve x^2+2x-120=0x2+2x−120=0x^2+2x-120=0 by completing the square?

1 Answer

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How do you solve $x^{2} + 2 x - 120 = 0$ $x^2+2x-120=0$ by completing the square?