Question

How to complete the square such that ax^2 + bx + c = a(x-p)^2 + q?

Answer 1

Do I do it like this: Please check my answers..

Work out the multiplication on the (x-p)^2 part.
y = a(x-p)^2 + q
y = a(x^2 - 2px + p^2) + q
y = ax^2 - 2apx + ap^2 + q

So now we have two formulas,
y = ax^2 + bx + c
y = ax^2 - 2apx + (ap^2 + q)

They're equal to each other for all values of x, so the coefficients must be the same.

The ax^2 is the same in both. a=a

bx = -2apx
so b = -2ap and
p = -b/2a

c = ap^2 + q so
q = c - ap^2
since p = -b/2a, you can also say
q = c - a(-b/2a)^2
q = c - a(b^2/4a^2)
q = c - b^2/4a

Answer 2

\[=a(x^2+\frac{b}{a}x+\frac{c}{a})\]\[=a[(x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}]\]\[ p=\frac{-b}{2a}\]\[q=-(\frac{b}{2a})^2+\frac{c}{a}\]