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

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

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Answer 1 · 2015-05-16T13:58:51

It's a rather strange way to solve it as it is easy to see that $x = 0$ $x=0$ and $x = - 1$ $x=-1$ are both solutions. However, here's how you would do it...

$0 = 2 x^{2} + 2 x = 2 {(x + \frac{1}{2})}^{2} - \frac{1}{2}$ $0 = 2x^2+2x = 2(x+1/2)^2 - 1/2$

Add 1/2 to both sides to get

$2 {(x + \frac{1}{2})}^{2} = \frac{1}{2}$ $2(x+1/2)^2 = 1/2$

Divide both sides by 2 to get

${(x + \frac{1}{2})}^{2} = \frac{1}{4}$ $(x+1/2)^2 = 1/4$

So

$x + \frac{1}{2} = \pm \sqrt{\frac{1}{4}} = \pm \frac{\sqrt{1}}{\sqrt{2^{2}}} = \pm \frac{1}{2}$ $x+1/2 = +-sqrt(1/4) = +-sqrt(1)/sqrt(2^2) = +-1/2$

Subtract 1/2 from both sides to get

$x = - \frac{1}{2} \pm \frac{1}{2} = \frac{- 1 \pm 1}{2}$ $x=-1/2+-1/2=(-1+-1)/2$

Answer 2 · 2015-05-17T04:03:17

If $2 x^{2} + 2 x = 0$ $2x^2 + 2x = 0$ , then $x^{2} + x = 0$ $x^2 + x = 0$ .

To complete this square, I need to square half of the coefficient of the 'x', which is 1/2.

${(\frac{1}{2})}^{2} = \frac{1}{4}$ $(1/2)^2=1/4$ , so $x^{2} + x + \frac{1}{4} = \frac{1}{4}$ $x^2 + x + 1/4 = 1/4$

${(x + \frac{1}{2})}^{2} = \frac{1}{4}$ $(x + 1/2)^2=1/4$

$∣ ∣ ∣ x + \frac{1}{2} ∣ ∣ ∣ = \sqrt{\frac{1}{4}}$ $|x + 1/2| = sqrt(1/4)$ (square roots can be positive or negative)

$∣ ∣ ∣ x + \frac{1}{2} ∣ ∣ ∣ = \frac{1}{2}$ $|x + 1/2| = 1/2$

$x = {- 1, 0}$ $x = {-1, 0}$

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

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