Year: 2010
Paper: 3
Question Number: 4
Course: LFM Stats And Pure
Section: Polynomials
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item
The number $\alpha$ is a common root of the
equations $x^2 +ax +b=0$ and $x^2+cx+d=0$
(that is, $\alpha$ satisfies both equations). Given that $a\ne c$, show that
\[
\alpha =- \frac{b-d}{a-c}\,.
\]
Hence, or otherwise, show that the equations have at least one
common root if and only if
\[
(b-d)^2 -a(b-d)(a-c) + b(a-c)^2 =0\,.
\]
Does this result still hold if the condition $a\ne c$ is not imposed?
\item Show that the equations
$x^2+ax+b=0$ and $x^3+(a+1)x^2+qx+r=0$
have at least one common root if and only if
\[
(b-r)^2-a(b-r)(a+b-q) +b(a+b-q)^2=0\,.
\]
Hence, or otherwise, find the values of $b$ for which the equations
$2x^2+5 x+2 b=0$ and $2x^3+7x^2+5x+1=0$
have at least one common root.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& 0 &= \alpha^2 + a \alpha + b \tag{1} \\
&& 0 &= \alpha^2 + c \alpha + d \tag{2} \\
\\
(1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\
\Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{$a\neq c$}
\end{align*}
($\Rightarrow$) Suppose they have a common root, then given we know it's form, we must have:
\begin{align*}
&& 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\
\Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2
\end{align*}
($\Leftarrow$) Suppose the equation holds, then
\begin{align*}
&& 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\
\Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\
\end{align*}
So $\alpha$ is a root of the first equation.
Considering $(1) - (2)$ we must have that $\alpha(a-c) +(b-d) = t$ (whatever the second equation is), but that value is clearly $0$, therefore $\alpha$ is a root of both equations.
If $a = c$ then the equation becomes $0 = (b-d)^2$, ie the two equations are the same, therefore they must have common roots!
\item
\begin{align*}
&& 0 &= x^2+ax+b \tag{1} \\
&& 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\
\\
(2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3}
\end{align*}
Therefore if the equations have a common root, $(1)$ and $(3)$ have a common root, ie $(b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0$ which is exactly our condition.
$a = \frac52, q = \frac52, r = \frac12$
\begin{align*}
&& 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\
&&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\
&&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\
&&&= (b-1)(4b^2-2b-1) \\
\Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}
\end{questionparts}
This was a popular question, though it was not generally well scored upon, with very few candidates earning full marks. Most began strongly, and finished by finding the values of b correctly. However, basic sign errors did prevent some from achieving the numerical pay-off. Part (ii) was, as expected, found trickier than part (i). Overall, the non-triviality of "if and only if" was rarely addressed as an issue in either part.