# Thread: Completing the square with fraction as coefficient of x^2

1. ## Completing the square with fraction as coefficient of x^2

Hi!

Complete the square for $\displaystyle \frac12x^2-x+3$

$\displaystyle \frac{-2}{2(1)}\,=\,1$

$\displaystyle \frac12(x^2 - 2x + 1 - 1) + 3$

$\displaystyle \frac12(x-1)^2 - 2$

2. ## Re: Completing the square with fraction as coefficient of x^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.

3. ## Re: Completing the square with fraction as coefficient of x^2

$\displaystyle \frac{-2}{2(1)}\,=\,1$

$\displaystyle \frac12(x^2 - 2x) + 3$

$\displaystyle \frac12\left[(x^2 - 2x + 1 - 1) + 3\right]$

$\displaystyle \frac12(x-1)^2 - 1 + 3$

$\displaystyle \frac12(x-1)^2 + 2$

4. ## Re: Completing the square with fraction as coefficient of x^2

I'm not giving any more help until you learn to follow the instructions you've been given.

5. ## Re: Completing the square with fraction as coefficient of x^2

My solution is what I understand from the instructions given.

Originally Posted by Prove It
The first thing to do is to take out $\displaystyle \frac12$ as a factor
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
then complete the square on everything left over
$\displaystyle \frac12(x^2 - 2x + 1 - 1 + 6)$

$\displaystyle \frac12\left[(x-1)^2 + 5\right]$

Originally Posted by Prove It
then multiply that factor back through.
$\displaystyle \frac12(x-1)^2 + \frac52$

6. ## Re: Completing the square with fraction as coefficient of x^2

And this is correct. Well done.