This is from the controversial GCSE question in the UKs recent exams. The orginal question is thus:

There are $n$ sweets in a bag. $6$ of the sweets are orange, the rest are yellow.

Hannah takes at random a sweet from the bag and eats it. She then takes another sweet at random from the bag and eats it.

The probability that hannah eats two orange sweets is 1/3

a) show that $n^2 - n - 90 = 0$

This in itself isn't too difficult:

$$ \frac{6}{n}\frac{5}{n-1} = \frac{30}{n^2 - n} = \frac{1}{3} \Rightarrow \frac{90}{n^2 -n} = 1 $$ so that $$ n^2 - n -90 = 0 $$ But, what I'm trying to figure out is the value of n, the only way to do this I have thought of is trial and error, quickly finding the answer to be 10. Is there a different option here?

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    $\begingroup$ Factorise the equation to give $(n-10)(n+9)=0$, resulting in $n=10$ as $n$ is positive. $\endgroup$ – hypergeometric Jun 9 '15 at 15:06
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    $\begingroup$ What about the question is controversial? $\endgroup$ – Daniel Fischer Jun 9 '15 at 15:07
  • $\begingroup$ If you assume what you are asked to show you can use the quadratic formula to find the roots of the equation or factorise as suggested in the comment above then test the conditions to check it works. $\endgroup$ – Warren Hill Jun 9 '15 at 15:08
  • $\begingroup$ Also: math.stackexchange.com/questions/1313076/… $\endgroup$ – MattAllegro Jun 9 '15 at 19:28
  • $\begingroup$ @DanielFischer It was a GCSE level question, there was no explanation based on the facts. It is somewhat advanced and was a bit here are some random facts, a b and c, and here's a question Z. Here is the news article: bbc.co.uk/news/education-33017299 $\endgroup$ – MrDobilina Jun 10 '15 at 9:37

You could factor $-90$. You need $a,b$ such that $ab = -90$ and $a+b = -1$. There aren't that many factors, so you can quickly find $a,b = -10,9$, so that $n^2 - n - 90 = (n-10)(n+9)$, which gets you to the solutions $n = 10, -9$, the second of which is obviously invalid.

Alternatively, if you don't notice the factoring, there's always the good old standby: the quadratic formula (assuming you know your squares!)

$$\begin{split} n &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{1 \pm \sqrt{361}}{2} \\ &= \frac{1 \pm 19}{2} \\ &= 10, -9 \end{split}$$

  • $\begingroup$ I do like this, had forgotten about the quadratic formula, been a while! $\endgroup$ – MrDobilina Jun 9 '15 at 15:18

I suppose use of quadratic formula can help(http://en.wikipedia.org/wiki/Quadratic_formula). Just substitute coefficients to expression $x=(-b-sqrt(b^2-4ac))/2a, y=(-b+sqrt(b^2-4ac))/2a$ where $a=1, b=-1, c=-90.$ Obviously, only positive solution makes sense, so you see n=10.


Ok write the quadratic expression as $$n^2-10n+9n-90=0$$ then factorise it to get $(n+9)(n-10)=0$.

$n$ can't be negative so the answer is only $10$.


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