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

That will come out ok: $\left[\left(x - \frac{59}{24}\right)^2\right] = \frac{59^2}{4\times 12^2} - 6$ - 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.

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.

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5 comments on “Ask Uncle Colin: An Absurd Quadratic

  • Christian Lawson-Perfect

    Eh? My first thought is that it’ll factorise as \( (4x – a)(3x – b) \), where \(a \cdot b = 72\) and \(3a + 4b = 59\). My first guess at factorising \(72\) is \(9 \times 8\), and I notice that \(a+b\) is odd, so it has to be \((4x-9)(3x-8)\).

    I don’t think it’s that unreasonable a quadratic, as far as GCSE quadratics go.

    I skipped a lot of trial and error by lucky choices there, which I think the setters will have been banking on.

    • Christian Lawson-Perfect

      Sorry, that’s \(3a+4b\) is odd.

      • Colin

        Why does it have to be $3x$ and $4x$ rather than $x$ and $12x$? (Or was that one of the lucky guesses?)

        • Christian Lawson-Perfect

          That was one of my lucky guesses. There were only three choices for the factorisation of \(12x^2\), so it wouldn’t have taken too long to check them all.

          I wonder if there’s something interesting deep within this question: lots of my attempted factorisations end up producing \(12x^2 – 60x +72\), which has a common factor of \(12\). You can rule these factorisations out before expanding them just by noticing that one half has a constant common factor, e.g. \( (12-24)(x-3) = 12(x-2)(x-3)\)

          So the given equation is “one away” from a simpler one. Can we use that fact to home in on the solution more quickly?

          • Christian Lawson-Perfect

            In fact, you can rule out \( (2x+a)(6x+b) \) immediately because there’s no way that’ll produce an odd \(x\) term.

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