1. ## Simple algebra

$x^2-8x+23=(x-p)^2+q$ 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 $-8=p^2-p^2$

What now?

2. Originally Posted by Mukilab
$x^2-8x+23=(x-p)^2+q$ 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 $-8=p^2-p^2$

What now?
No, you should complete the square on the left hand side to get it into the form on the right, then find p and q by comparing coefficients

To get you started $x^2+bx = \left(x+\frac{b}{2}\right)^2 - \frac{b^2}{4}
$

3. Originally Posted by e^(i*pi)
No, you should complete the square on the left hand side to get it into the form on the right, then find p and q by comparing coefficients

To get you started $x^2+bx = \left(x+\frac{b}{2}\right)^2 - \frac{b^2}{4}
$
Do not know what you mean by
" complete the square on the left "
or
"comparing coefficients"

4. Originally Posted by Mukilab
Do not know what you mean by
" complete the square on the left "
or
"comparing coefficients"
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