How do you complete the square for $x^2 -3x$ ?

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Answer 1 · 2015-05-24T13:55:22

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

To see this, let's multiply out...

$(x-3/2)^2 = (x-3/2)(x-3/2)$

$= x(x-3/2)-3/2(x-3/2)$

$=x^2-3/2x-3/2x+9/4 = x^2-3x+9/4$

Subtract $9/4$ from both sides to get $x^2-3x = (x-3/2)^2 - 9/4$

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

In your case we have $a=1$ , $b=-3$ and $c=0$ .

How do you complete the square for x^2 -3xx^2 -3x?