
quadratic problem
$\displaystyle x^2  14x + a = (x + b)^2$
for all values of x, ive been asked to find the value of a and the value of b
im not sure, but i think the solution to this may lie in 'completing the square'.
$\displaystyle (x7)^2  49 + a = (x + \frac{2b}{2})^2  \frac{4b^2}{4} + b^2$
i think i was going round in circles (Happy)
$\displaystyle x^2 14x + 49  49 + a = x^2 + 2bx + \frac{4b^2}{4}  \frac{4b^2}{4} + b^2$ expanding the above
$\displaystyle x^2 14x + a = x^2 + 2bx + b^2$ simplify
$\displaystyle 14x + a=2bx + b^2$ and cancel

$\displaystyle x^214x+a=x^2+2bx+b^2$
Then, $\displaystyle \left\{\begin{array}{ll}2b=14\\b^2=a\end{array}\right.$
Now, solve the system.

thanks for that, should have stepped back a bit before trying it (Clapping)