Solve the quadratic equation : $x^2 + 2\sqrt{2}x - 6 = 0$ for $x$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer
Here, $a = 1,\ b = 2\sqrt{2},\ c = -6$.
Discriminant $= b^2 - 4ac = (2\sqrt{2})^2 - 4(1)(-6) = 8 + 24 = 32 > 0$
Using the quadratic formula:
$$x = \frac{-2\sqrt{2} \pm \sqrt{32}}{2} = \frac{-2\sqrt{2} \pm 4\sqrt{2}}{2}$$
$$x = \frac{-2\sqrt{2} + 4\sqrt{2}}{2} = \sqrt{2} \quad \text{or} \quad x = \frac{-2\sqrt{2} - 4\sqrt{2}}{2} = -3\sqrt{2}$$
Therefore, $x = \sqrt{2}$ or $x = -3\sqrt{2}$.
Source: Chapter 4, Section 4.4
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Explanation
- Identify $a$, $b$, $c$ first, then find the discriminant to confirm real roots exist.
- Note $\sqrt{32} = 4\sqrt{2}$ — simplifying surds is a common error point.
- Show both roots clearly; examiners award 1 mark for correct formula/discriminant and 1 mark for both correct roots.