How do I complete the square?

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Answer 1 · 2015-10-23T20:16:56

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

The secret is that $b/(2a)$ bit

Suppose you are given a quadratic equation to solve:

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

..which is in the form..

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

$b/(2a) = -3/4$

So we find:

$2(x-3/4)^2 = 2(x^2-(2*x*3/4)+(3/4)^2)$

$=2(x^2-(3x)/2+9/16)$

$=2x^2-3x+9/8$

So:

$2(x-3/4)^2-25/8 = 2(x-3/4)^2-9/8-2$

$=2x^2-3x+9/8-9/8-2$

$=2x^2-3x-2$

So:

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

turns into:

$2(x-3/4)^2-25/8 = 0$

Hence:

$(x-3/4)^2 = 25/16$

So:

$x-3/4 = +-sqrt(25/16) = +-5/4$

and

$x = 3/4+-5/4$