$\displaystyle x^2-8x+23=(x-p)^2+q$ for all values of x

Find p and q, does this mean I can just use x as anything such as 0?

if so using simultaneous equations I got $\displaystyle -8=p^2-p^2$

What now?

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- May 20th 2010, 12:10 PMMukilabSimple algebra
$\displaystyle x^2-8x+23=(x-p)^2+q$ for all values of x

Find p and q, does this mean I can just use x as anything such as 0?

if so using simultaneous equations I got $\displaystyle -8=p^2-p^2$

What now? - May 20th 2010, 12:28 PMe^(i*pi)
- May 20th 2010, 12:50 PMMukilab
- May 20th 2010, 01:51 PMe^(i*pi)
Completing the square is covered in post 3 of this topic: http://www.mathhelpforum.com/math-he...tic-curve.html

Comparing coefficients says that two equations are equal if and only if the coefficients on each side match.

For example given that ax^2+bx+c = x^2 - 12x+1 comparing coefficients tells you that a = 1, b=-12 and c = 1 - May 21st 2010, 08:30 AMMukilab
I have several problems with 'completing the square', please read that thread.

Thanks for telling me about comparing the coefficient :)