$\displaystyle x^2 - 14x + a = (x + b)^2$

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'.

$\displaystyle (x-7)^2 - 49 + a = (x + \frac{2b}{2})^2 - \frac{4b^2}{4} + b^2$

i think i was going round in circles

$\displaystyle x^2 -14x + 49 - 49 + a = x^2 + 2bx + \frac{4b^2}{4} - \frac{4b^2}{4} + b^2$ expanding the above

$\displaystyle x^2 -14x + a = x^2 + 2bx + b^2$ simplify

$\displaystyle -14x + a=2bx + b^2$ and cancel