• February 26th 2009, 01:14 AM
sammy28
$x^2 - 14x + a = (x + b)^2$

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'.

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

i think i was going round in circles (Happy)

$x^2 -14x + 49 - 49 + a = x^2 + 2bx + \frac{4b^2}{4} - \frac{4b^2}{4} + b^2$ expanding the above
$x^2 -14x + a = x^2 + 2bx + b^2$ simplify
$-14x + a=2bx + b^2$ and cancel
• February 26th 2009, 01:31 AM
red_dog
$x^2-14x+a=x^2+2bx+b^2$

Then, $\left\{\begin{array}{ll}2b=-14\\b^2=a\end{array}\right.$

Now, solve the system.
• February 26th 2009, 01:55 AM
sammy28
thanks for that, should have stepped back a bit before trying it (Clapping)