hi all,

the question posed is

Given that $\displaystyle x^2-14x+a = (x+b)^2$ for all values of $\displaystyle x$, find the value of $\displaystyle a$ and the value of $\displaystyle b$.

is this correct?

since

$\displaystyle (x+A)^2=x^2+2Ax+A^2$

$\displaystyle x^2+2Ax=(x+A)^2-A^2$

so

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

and solving for b

$\displaystyle b=\frac{-14x}{2x}=-7$

so

$\displaystyle x^2-14x+a=(x-7)^2$

and expanding the RHS

$\displaystyle a=49$