Find the zeroes of the quadratic polynomial $x^2 - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
Finding zeroes:
Using the identity $a^2 - b^2 = (a-b)(a+b)$:
$$x^2 - 15 = \left(x - \sqrt{15}\right)\left(x + \sqrt{15}\right)$$
So, $x^2 - 15 = 0$ when $x = \sqrt{15}$ or $x = -\sqrt{15}$.
The zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Verification:
Here $a = 1$, $b = 0$, $c = -15$.
$$\text{Sum of zeroes} = \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a} \checkmark$$
$$\text{Product of zeroes} = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a} \checkmark$$
Hence, the relationship is verified.
Source: Chapter 2, Section 2.3
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Explanation
- This is modelled on Example 3 of the textbook (which uses $x^2 - 3$); the same method applies to $x^2 - 15$.
- Examiners award marks for: (1) correct factorisation/zeroes, (2) correct sum verification, (3) correct product verification. All three steps must be shown explicitly.
- Write $\frac{-b}{a}$ and $\frac{c}{a}$ with actual coefficient values substituted — don't just state the formula.
- The coefficient of $x$ is 0 (not absent), so $b = 0$; state this clearly to avoid losing marks.