How do you solve $5 x^{2} + 20 x - 120 = 0$ $5x^2 + 20x - 120 = 0$ using completing the square?

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How do you solve $5 x^{2} + 20 x - 120 = 0$ $5x^2 + 20x - 120 = 0$ using completing the square?

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Answer 1 · 2015-07-03T12:27:04

Logical first step would be to divide everything by $5$ $5$

Explanation:

$\to x^{2} + 4 x - 24 = 0$ $->x^2+4x-24=0$

Since ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2=x^2 +2ax+a^2$
We take half of the coefficient of $x$ $x$ , and square it:
$x^{2} + 4 x + 4 = {(x + 2)}^{2}$ $x^2+4x+4=(x+2)^2$

We have to balance the $4$ $4$ we used with the $- 24$ $-24$
So the equation becomes:
$x^{2} + 4 x + 4 - 28 = 0 \to$ $x^2+4x+4-28=0->$ add 28:
${(x + 2)}^{2} = 28$ $(x+2)^2=28$

So $x + 2 = \sqrt{28} = 2 \sqrt{7} \to x = - 2 + 2 \sqrt{7}$ $x+2=sqrt28=2sqrt7->x=-2+2sqrt7$
Or $x + 2 = - \sqrt{28} = - 2 \sqrt{7} \to x = - 2 - 2 \sqrt{7}$ $x+2=-sqrt28=-2sqrt7->x=-2-2sqrt7$

(sometimes written as $x_{1, 2} = - 2 \pm 2 \sqrt{7}$ $x_(1,2)=-2+-2sqrt7$ )

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How do you solve $5 x^{2} + 20 x - 120 = 0$ $5x^2 + 20x - 120 = 0$ using completing the square?

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How do you solve $5 x^{2} + 20 x - 120 = 0$ $5x^2 + 20x - 120 = 0$ using completing the square?