How do you solve $0 = x^{2} - 3 x - 6$ $0=x^2-3x-6$ by completing the square?

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Answer 1 · 2015-05-17T07:33:03

$0 = x^{2} - 3 x - 6 = (x^{2} - 3 x + \frac{9}{4}) - \frac{9}{4} - 6$ $0 = x^2-3x-6 = (x^2-3x+9/4)-9/4-6$

$= {(x - \frac{3}{2})}^{2} - \frac{33}{4}$ $=(x-3/2)^2-33/4$

In general:

$a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))$

Adding $\frac{33}{4}$ $33/4$ to both sides we get

${(x - \frac{3}{2})}^{2} = \frac{33}{4}$ $(x-3/2)^2 = 33/4$

So $x - \frac{3}{2} = \pm \sqrt{\frac{33}{4}} = \pm \frac{\sqrt{33}}{2}$ $x-3/2 = +-sqrt(33/4) = +-sqrt(33)/2$

Adding $\frac{3}{2}$ $3/2$ to both sides we get

$x = \frac{3}{2} \pm \frac{\sqrt{33}}{2} = \frac{3 \pm \sqrt{33}}{2}$ $x = 3/2+-sqrt(33)/2 = (3+-sqrt(33))/2$

How do you solve 0=x2−3x−60=x^2-3x-6 by completing the square?