How do you solve by completing the square: $- 2 x^{2} - 3 x - 3 = 0$ $-2x^2-3x-3=0$ ?

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How do you solve by completing the square: $- 2 x^{2} - 3 x - 3 = 0$ $-2x^2-3x-3=0$ ?

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Answer 1 · 2015-03-31T11:25:30

$- 2 x^{2} - 3 x - 3 = 0$ $-2x^2-3x-3 = 0$

Divide all terms of both sides of the equation by $(- 2)$ $(-2)$ so we are working with an expression that begins simply with $x^{2}$ $x^2$

$x^{2} + \frac{3}{2} x + \frac{3}{2} = 0$ $x^2 +3/2x + 3/2 = 0$

then move the constant $(- \frac{3}{2})$ $(-3/2)$ off the left-side of the equation by subtracting $\frac{3}{2}$ $3/2$ from both sides
$x^{2} + \frac{3}{2} x = - \frac{3}{2}$ $x^2+3/2x = -3/2$

To "complete the square" we need something of the form:
${(x + a)}^{2}$ $(x+a)^2$
which equals $x^{2} + 2 a x + a^{2}$ $x^2+2ax +a^2$

Since we have computed the coefficient of $x$ $x$ to be $\frac{3}{2}$ $3/2$
$\to a = \frac{3}{4}$ $rarr a = 3/4$
and $a^{2} = \frac{9}{4}$ $a^2 = 9/4$

Add $\frac{9}{4}$ $9/4$ to both sides of the equation
$x^{2} + \frac{3}{2} x + \frac{9}{4} = \frac{9}{4} - \frac{3}{2}$ $x^2+3/2x+9/4 = 9/4 - 3/2$

Rewrite the left-side as a square and simplify the right-side
${(x + \frac{3}{2})}^{2} = \frac{3}{4}$ $(x+3/2)^2 = 3/4$

Take the square root of both sides
$x + \frac{3}{2} = \pm \frac{\sqrt{3}}{2}$ $x+3/2 = +-sqrt(3)/2$

$x = - \frac{3 + \sqrt{3}}{2}$ $x = -(3+sqrt(3))/2$
or
$x = - \frac{3 - \sqrt{3}}{2} = \frac{\sqrt{3} - 3}{2}$ $x = -(3-sqrt(3))/2 = (sqrt(3)-3)/2$

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How do you solve by completing the square: $- 2 x^{2} - 3 x - 3 = 0$ $-2x^2-3x-3=0$ ?

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