Year: 2017
Paper: 3
Question Number: 3
Course: UFM Pure
Section: Roots of polynomials
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the
quartic equation
\[
x^4 +px^3 +qx^2 +r x +s =0
\,.
\]
You are given that, for any such equation, $\,\alpha \beta + \gamma\delta\,$, $\alpha\gamma+\beta\delta\,$ and $\,\alpha \delta + \beta\gamma\,$ satisfy a cubic equation of the form
\[
y^3+Ay^2+
(pr-4s)y+ (4qs-p^2s -r^2)
=0
\,.
\]
Determine $A$.
Now consider the quartic equation given by $p=0\,$, $q= 3\,$, $r=-6\,$ and $s=10\,$.
\begin{questionparts}
\item
Find the value of $\alpha\beta + \gamma \delta$, given that it is the largest root of the corresponding cubic equation.
\item
Hence, using the values of $q$ and $s$, find the value of $(\alpha +\beta)(\gamma+\delta)\,$
and the value of $\alpha\beta$ given that $\alpha\beta >\gamma\delta\,$.
\item
Using these results, and the values of $p$ and $r$, solve the quartic equation.
\end{questionparts}\begin{align*}
A &= -(\alpha \beta + \gamma\delta + \alpha\gamma+\beta\delta+\alpha \delta + \beta\gamma) \\
&= -q
\end{align*}
\begin{questionparts}
\item The corresponding cubic equation is:
\begin{align*}
&& 0 &= y^3 - 3y^2-40y+(120-36) \\
&&&= y^3 -3y^2 - 40y + 84 \\
&&&= (y-7)(y-2)(y+6)
\end{align*}
Therefore $\alpha\beta + \gamma \delta = 7$
\item
\begin{align*}(\alpha+\beta)(\gamma+\delta) &= \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\
&= 3 -(\alpha\beta + \gamma\delta) \\
&=3-7 = -4
\end{align*}
Let $\alpha\beta$ and $\gamma\delta$ be the roots of a quadratic; then the quadratic will be $t^2-7t+10 = 0 \Rightarrow t = 2,5$ so $\alpha\beta = 5$
\item $\alpha\beta = 5, \gamma\delta = 2$
Consider the quadratic with roots $\alpha+\beta$ and $\gamma+\delta$, then
$t^2-4 = 0 \Rightarrow t = \pm 2$.
Suppose $\alpha+\beta = 2, \gamma+\delta=-2$ then
$\alpha, \beta = 1 \pm 2i, \gamma,\delta = -1 \pm i$
$\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = -6 \neq 6$
Suppose $\alpha+\beta = -2, \gamma+\delta=2$ then
$\alpha, \beta = -1 \pm 2i, \gamma,\delta = 1 \pm i$
$\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = 6$, therefore these are there roots. (In some order):
$1 \pm i, -1 \pm 2i$
\end{questionparts}
The second most popular question at just over 70%, the success rate was about half marks in common with a number of other questions, with the majority earning either 16 and above, or 4 and below. A common mistake was omitting the minus sign in the first step to obtain which resulted in candidates being unable to progress further. If the cubic equation was correctly found, then candidates tended to score all the marks as far as part (iii). A few candidates obtaining the correct results in (iii) then stated that the answers could not be complex, which was, of course, false.