| Date | November 2009 | Marks available | 7 | Reference code | 09N.2.hl.TZ0.7 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
Consider the equation \({z^3} + a{z^2} + bz + c = 0\) , where \(a\) , \(b\), \(c \in \mathbb{R}\) . The points in the Argand diagram representing the three roots of the equation form the vertices of a triangle whose area is \(9\). Given that one root is \( - 1 + 3{\text{i}}\) , find
(a) the other two roots;
(b) \(a\) , \(b\) and \(c\) .
Markscheme
(a) one root is \(- 1 - 3{\text{i}}\) A1
distance between roots is \(6\), implies height is \(3\) (M1)A1
EITHER
\( - 1 + 3 = 2 \Rightarrow \) third root is \(2\) A1
OR
\( - 1 - 3 = - 4 \Rightarrow \) third root is \(4\) A1
(b) EITHER
\(z - \left( { - 1 + 3{\text{i}}} \right)\) \(z - \left( { - 1 - 3{\text{i}}} \right)\) \(\left( {z - 2} \right) = 0\) M1
\( \Rightarrow \left( {{z^2} + 2z + 10} \right)\left( {z - 2} \right) = 0\) (A1)
\({{z^2} + 6z - 20 = 0}\) A1
\(a = 0\), \(b = 6\) and \(c = 20\)
OR
\(z - \left( { - 1 + 3{\text{i}}} \right)\) \(z - \left( { - 1 - 3{\text{i}}} \right)\) \(\left( {z + 4} \right) = 0\) M1
\( \Rightarrow \left( {{z^2} + 2z + 10} \right)\left( {z + 4} \right) = 0\) (A1)
\({{z^2} + 6{z^2} + 18z + 40 = 0}\) A1
\(a = 6\), \(b = 18\) and \(c = 40\)
[7 marks]
Examiners report
Most students were able to state the conjugate root, but many were unable to take the question further. Of those that then recognised the method, the question was well answered.
