Dear Uncle Colin,

I was asked to complete the square on $f(x) = 2x^2 + 13x + 20$. I started by halving everything, which makes it cleaner, but the solution manual disagrees. What gives?

- Have Always Loathed Functions

Hi, HALF, and thanks for your message!

Dividing a quadratic by the $x^2$ coefficient -- here, 2 -- to make the numbers easier is often a good strategy, but you have to be careful. In particular, here you have an equation: $f(x) = 2x^2 + 13x + 20$ - and you can't just halve one side of it unless you halve the other.

So, you can tackle it that way, *as long as you account for the two*.

You can do $\frac{1}{2}f(x) = x^2 + \frac{13}{2}x + 10$, then:

$\frac{1}{2}f(x) = \br{x+\frac{13}{4}}^2 - \br{\frac{13}{4}}^2 + 10$

$\frac{1}{2}f(x) = \br{x+\frac{13}{4}}^2 - \frac{9}{16}$

$f(x) = 2\br{x+\frac{13}{4}}^2 - \frac{9}{8}$

### Another explanation

When you worked through, presumably you got an answer of $\br{x+\frac{13}{4}}^2 - \frac{9}{16}$. You can see that this isn't the same as $f(x)$ by substituting a value into both - for example, $0$.

In the original function, $f(0)=20$. In the halved answer, $\br{\frac{13}{4}}^2 - \frac{9}{16} = 10$, which is not the same.

Since the idea of completing the square here is to write the same function in a different way, it's no good to have different answers for the same value! The two things aren't equivalent, so something has gone wrong.

### When you *can* halve it

You can halve the expression without any penalty *if you know it's equal to zero* - for example, if you're trying to find the roots. In that case, halving the other side of the equation doesn't change it, and everything is hunky-dory.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.