Ask Uncle Colin: An Absurd Quadratic

Dear Uncle Colin,

In a recent exam, I was invited to solve $12x^2-59x+72=0$ without a calculator. Is that a reasonable thing to ask?

Very Irate EdExcel-Taught Examinee

Hi, VIETE, and I don’t blame you for being cross - in a non-calculator exam, I’m not sure that really tests the Supposedly Important Skills you’re meant to have.

That said, it’s not impossible.

Method 1: using the formula

This is a bit brutal, but it can be done: if $a=12$, $b=-59$ and $c=72$, you have $\frac{59\pm \sqrt{(-59{)}^2-4(12)(72)}}{2(12)}$. None of that is especially nice, but ${59}^2=3481$, which you could get by expanding $(60-1{)}^2$; $4\times 12\times 72=48\times 72=(60-12)(60+12)=3600-144=3456$.

The difference between those is 25, so the big square root becomes 5. Your answers are $\frac{59\pm 5}{24}$, or $\frac{64}{24}=\frac{8}{3}$ and $\frac{54}{24}=\frac{9}{4}$.

Method 2: factorising (1)

You can also play the “magic numbers” game I’ve talked about elsewhere, and find factors of $12\times 72$ that sum to $-59$. Note that I don’t really care what $12\times 72$ is - I’m just going to work with the factors about until I find a pair that works.

Some things I notice: both of the magic numbers must be negative; also, one must be even and one odd (otherwise their sum would be even).

And whatever $12\times 72$ is, it doesn’t have all that many odd factors. Its prime factorisation is $2^5\times 3^3$, so its only negative odd factors are -1, -3, -9 and -27. Of those, $(-27)+(-32)$ gives the required $-59$.

Splitting the original expression up as $12x^2-27x-32x+72$, I get $3x(4x-9)-8(4x-9)$ or $(3x-8)(4x-9)$ as before.

Method 3: factorising (2)

That method did rather depend on spotting something nice about the factors. An alternative approach is a trial and improvement method to see which factor pairs are too close together or too far apart.

For example, we know that $-12\times -72$ would give the necessary number - however, $(-12)+(-72)=-84$, which means that pair is too far apart. So, let’s try doubling the -12 and halving the -72:

$(-24)+(-36)=-60$, which is close, but still too far apart - so our factors need to be between -24 and -36. At least one of them has to be a multiple of 3, so our candidates are -27, -30 and -33; -30 can’t work (there’s no factor of 5) and neither can -33 (no factor of 11), so -27 is our only shot. A little work (multiplying the first by $\frac{9}{8}$ and the second by $\frac{8}{9}$ gives the same -27 and -32 as before).

Method 4: completing the square.

Seriously? I mean… you can do it, but I’m not sure you’d want to.

You end up with something like $12\left[{(x-\frac{59}{24})}^2\right]-\frac{{59}^2}{48}+72=0$.

That will come out ok: $\left[{(x-\frac{59}{24})}^2\right]=\frac{{59}^2}{4\times {12}^2}-6$ - but this boils down to the same arithmetic in method 1.

So, VIETE: it’s possible to do, but I don’t think it’s an especially good thing to be asked to do under pressure in an exam.

Hope that helps!

C

* Edited 2017-08-16 to fix a LaTeX error.