How do I complete the square?

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How do I complete the square?

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Answer 1 · 2015-10-23T20:16:56

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

The secret is that $\frac{b}{2 a}$ $b/(2a)$ bit

Explanation:

Suppose you are given a quadratic equation to solve:

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

..which is in the form..

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ with $a = 2$ $a = 2$ , $b = - 3$ $b=-3$ and $c = - 2$ $c=-2$

$\frac{b}{2 a} = - \frac{3}{4}$ $b/(2a) = -3/4$

So we find:

$2 {(x - \frac{3}{4})}^{2} = 2 (x^{2} - (2 \cdot x \cdot \frac{3}{4}) + {(\frac{3}{4})}^{2})$ $2(x-3/4)^2 = 2(x^2-(2*x*3/4)+(3/4)^2)$

$= 2 (x^{2} - \frac{3 x}{2} + \frac{9}{16})$ $=2(x^2-(3x)/2+9/16)$

$= 2 x^{2} - 3 x + \frac{9}{8}$ $=2x^2-3x+9/8$

So:

$2 {(x - \frac{3}{4})}^{2} - \frac{25}{8} = 2 {(x - \frac{3}{4})}^{2} - \frac{9}{8} - 2$ $2(x-3/4)^2-25/8 = 2(x-3/4)^2-9/8-2$

$= 2 x^{2} - 3 x + \frac{9}{8} - \frac{9}{8} - 2$ $=2x^2-3x+9/8-9/8-2$

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

So:

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

turns into:

$2 {(x - \frac{3}{4})}^{2} - \frac{25}{8} = 0$ $2(x-3/4)^2-25/8 = 0$

Hence:

${(x - \frac{3}{4})}^{2} = \frac{25}{16}$ $(x-3/4)^2 = 25/16$

So:

$x - \frac{3}{4} = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}$ $x-3/4 = +-sqrt(25/16) = +-5/4$

and

$x = \frac{3}{4} \pm \frac{5}{4}$ $x = 3/4+-5/4$

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How do I complete the square?

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