2021 H2 Maths Paper 2, Question 1

As $a$ and $b$ are real, the complex roots of $x^{3} + 2x^{2} + ax + b = 0$ occur in conjugate pairs. Since $1 + \frac{1}{2}i$ is a root, its conjugate $1 - \frac{1}{2}i$ is also a root.

Let the third root be $r$, which must be real. By Vieta’s formula, the sum of the roots equals $- 2$ (the negative of the coefficient of $x^{2}$):

To check your answer: the three roots $1 + \frac{1}{2}i$, $1 - \frac{1}{2}i$, $- 4$ sum to $2 + ( - 4) = - 2$, as required ✓

By Vieta’s formula, the product of the roots equals $- b$:

Since $\left( 1 + \frac{1}{2}i \right)\left( 1 - \frac{1}{2}i \right) = 1^{2} - \left( \frac{1}{2}i \right)^{2} = 1 + \frac{1}{4} = \frac{5}{4}$,

Also by Vieta’s formula, the sum of the products of the roots taken two at a time equals $a$:

To check your answer: expanding $\left( x - 1 - \frac{1}{2}i \right)\left( x - 1 + \frac{1}{2}i \right)(x + 4)$ directly gives $x^{3} + 2x^{2} - \frac{27}{4}x + 5$, matching $a = - \frac{27}{4}$ and $b = 5$ ✓