I am asked to solve for $x$ by completing the square: $x^2-9x-22$
I arrive at a situation where I try to factor $x^2-9x+9$ as a perfect square but I'm unable to see how.
$x^2-9x-22$
$x^2-9x=22$
Take $\frac{1}{2}$ of $b$ then square it:
$\frac{1}{2}*\frac{-9}{1}=\frac{-9}{3}=-3$
$-3^2=9$
Add this to both sides:
$x^2-9x+9=31$
That's as far as I get. I cannot see how to factor the left hand side as a perfect square?
I think I was supposed to use the number I arrived at when taking $\frac{1}{2}$ of $b$ before squaring which was $-3$. But $(x-3)^2$ does not equal
$x^2-9x+9$?
Best Answer
Here is the idea. When completing the square you have $$ (ax+b)^2 = a^2 x^2 + 2abx + b^2, $$ which in your case must fit $x^2-9x+9$. From the leading coefficient, it's easy to see we must have $a=1$, so we are fitting $$ (x+b)^2 = x^2+2bx+b^2 $$ to $x^2-9x+9$.
It's easy to see from the linear term that $2b = -9$ so $b= -9/2$ and you are fitting $$ (x-9/2)^2 = x^2-9x +(9/2)^2. $$ Can you now complete the problem?
UPDATE
From your comment, you are seeking not to complete the square but to factor the polynomial. the most direct way is to find the roots using the quadratic formula, to get $$ r_\pm = \frac{-(-9)\pm \sqrt{(-9)^2-4 \cdot 1 \cdot (-22)}}{2\cdot 1} = \frac{9 \pm \sqrt{169}}{2} = \frac{9 \pm 13}{2} = \{11,-2\} $$ Therefore your polynomial must factor as $$ x^2-9x-22 = (x-r_+)(x-r_-) = (x-11)(x+2). $$