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