Without solving the equation, find the value of

(i)     $\alpha  + \beta$;

(ii)     $\alpha \beta$.

Another quadratic equation ${x^2} + px + q = 0,{\text{ }}p,{\text{ }}q \in \mathbb{Z}$ has roots $\frac{2}{\alpha }$ and $\frac{2}{\beta }$.

Find the value of $p$ and the value of $q$.

using the formulae for the sum and product of roots:

(i)     $\alpha  + \beta  = 4$     A1

(ii)     $\alpha \beta  = \frac{1}{2}$     A1

Note:     Award A0A0 if the above results are obtained by solving the original equation (except for the purpose of checking).

[2 marks]

METHOD 1

required quadratic is of the form ${x^2} - \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)x + \left( {\frac{2}{\alpha }} \right)\left( {\frac{2}{\beta }} \right)$     (M1)

$q = \frac{4}{{\alpha \beta }}$

$q = 8$     A1

$p =  - \left( {\frac{2}{\alpha } + \frac{2}{\beta }} \right)$

$=  - \frac{{2(\alpha  + \beta )}}{{\alpha \beta }}$     M1

$=  - \frac{{2 \times 4}}{{\frac{1}{2}}}$

$p =  - 16$     A1

Note:     Accept the use of exact roots

METHOD 2

replacing $x$ with $\frac{2}{x}$     M1

$2{\left( {\frac{2}{x}} \right)^2} - 8\left( {\frac{2}{x}} \right) + 1 = 0$

$\frac{8}{{{x^2}}} - \frac{{16}}{x} + 1 = 0$     (A1)

${x^2} - 16x + 8 = 0$

$p =  - 16$ and $q = 8$     A1A1

Note:     Award A1A0 for ${x^2} - 16x + 8 = 0$ ie, if $p =  - 16$ and $q = 8$ are not explicitly stated.

[4 marks]

Total [6 marks]

Most candidates obtained full marks.

Many candidates obtained full marks, but some responses were inefficiently expressed. A very small minority attempted to use the exact roots, usually unsuccessfully.