AS 91577 · 2025 paper · a remainder, a quadratic in k, a discriminant proof, a cube root, and a hyperbola
Question 1(c) Merit
Practises Polynomial equations with complex roots
Five separate parts: a remainder, a quadratic solved in terms of a constant, a proof that a quadratic always has real roots, the cube roots of an imaginary number, and a locus that turns out to be a hyperbola.
Prove that the equation (kx)2 = 3 − x/k has real roots for all real values of k, where k ≠ 0.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 The discriminant needs the form ax2 + bx + c = 0. Rearrange.
k2x2 + x/k − 3 = 0, so a = k2, b = 1/k and c = −3. Since k ≠ 0, a is not zero, so it really is a quadratic. -
2 Work out b2 − 4ac.
b2 − 4ac = (1/k)2 − 4(k2)(−3) = 1/k2 + 12k2
If you clear the fraction first (multiply through by k), you get k3x2 + x − 3k = 0 and b2 − 4ac = 1 + 12k4. The argument below works for either. -
3 Why can that never be negative, or zero, whatever real k is?
For any real k ≠ 0, k2 > 0. So 1/k2 and 12k2 are both positive, and so is their sum. -
4 What does a positive discriminant tell you?
Two distinct real roots. So for every real k ≠ 0 the equation has real roots, which is what had to be proved.
What the marker wanted
- Achieved The discriminant worked out correctly, but with no reason given for why it is always positive.
- Merit The proof in full: the discriminant, and why it is positive for every k (a non-zero square is positive), so the roots are real.
In our words, from NZQA’s 2025 marking schedule.
✕Common mistake
Trying values of k. NZQA’s 2025 report says many students misread what a proof asks for and set out to find which k values make the roots real. A proof has to cover every k at once, which is why the reason the discriminant is positive is the Merit step.How did it go?
Got it moves it further out. Review brings it back tomorrow.