Hi!
Complete the square for $\displaystyle \frac12x^2-x+3$
$\displaystyle \frac{-2}{2(1)}\,=\,1$
$\displaystyle \frac12(x^2 - 2x + 1 - 1) + 3$
$\displaystyle \frac12(x-1)^2 - 2$
You make life difficult on yourself if you skip steps. The first thing to do is to take out $\displaystyle \displaystyle \begin{align*} \frac{1}{2} \end{align*}$ as a factor, then complete the square on everything left over, then multiply that factor back through.
$\displaystyle \frac{-2}{2(1)}\,=\,1$
$\displaystyle \frac12(x^2 - 2x) + 3$
$\displaystyle \frac12\left[(x^2 - 2x + 1 - 1) + 3\right]$
$\displaystyle \frac12(x-1)^2 - 1 + 3$
$\displaystyle \frac12(x-1)^2 + 2$
My solution is what I understand from the instructions given.
Factoring $\displaystyle \frac12$ from the $\displaystyle x^2$ and $\displaystyle x$ terms: $\displaystyle \frac12\left[(x^2 - 2x + 6)\right]$Originally Posted by Prove It
$\displaystyle \frac12(x^2 - 2x + 1 - 1 + 6)$Originally Posted by Prove It
$\displaystyle \frac12\left[(x-1)^2 + 5\right]$
$\displaystyle \frac12(x-1)^2 + \frac52$Originally Posted by Prove It