If $x = -2$ is the common solution of quadratic equations $ax^2 + x - 3a = 0$ and $x^2 + bx + b = 0$, then find the value of $a^2 b$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer
Since $x = -2$ is a root of $ax^2 + x - 3a = 0$:
$$a(-2)^2 + (-2) - 3a = 0$$
$$4a - 2 - 3a = 0 \implies a = 2$$
Since $x = -2$ is a root of $x^2 + bx + b = 0$:
$$(-2)^2 + b(-2) + b = 0$$
$$4 - 2b + b = 0 \implies b = 4$$
Therefore, $a^2b = (2)^2 \times 4 = \mathbf{16}$
Source: Chapter 4, Section 4.3
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Explanation
- A root satisfies the equation, so substitute $x = -2$ into each equation separately and solve for the unknown constant.
- Find $a$ from the first equation, $b$ from the second, then compute $a^2b$.
- Examiners award 1 mark for finding both $a$ and $b$ correctly, and 1 mark for the final value. Show substitution steps clearly.