← Complex numbers: past papers

AS 91577 · 2025 paper · a remainder, a quadratic in k, a discriminant proof, a cube root, and a hyperbola

Question 1(b) Achieved

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.

Solve the equation x2 − 6kx = k2 for x. Simplify your answer as far as possible, giving it in terms of k.

The 2025 paper, page 2 ↗ Marking schedule, page 1 ↗

Work it out

Try it on paper first. Stuck? Take one hint at a time.

  1. 1 Move everything to one side. What kind of equation is it in x?

    x2 − 6kx − k2 = 0: a quadratic in x, with a = 1, b = −6k and c = −k2.

  2. 2 Use the quadratic formula, treating k as if it were a number.

    x = (6k ± √(36k2 + 4k2))/2 = (6k ± √40k2)/2

  3. 3 Simplify the root. What is √40k2?

    √40k2 = √(4 × 10 × k2) = 2√10 |k|. The ± in front already covers both signs, so |k| can be written as k.
    x = (6k ± 2√10 k)/2 = 3k ± √10 k

  4. 4 Can you check it another way?

    Complete the square: x2 − 6kx + 9k2 = 10k2, so (x − 3k)2 = 10k2 and x − 3k = ±√10 k. The same two answers.
Answer x = 3k + √10 k or x = 3k − √10 k, which is x = k(3 ± √10).

What the marker wanted

  • Achieved Both solutions, simplified fully: the root of 40k2 taken out as 2√10 k.

In our words, from NZQA’s 2025 marking schedule.

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Common mistake

Not getting started. NZQA’s 2025 report says many students could not solve a quadratic with a letter in its coefficients, managing it with neither the formula nor completing the square, and puts that down to leaning on a graphics calculator, which cannot give x in terms of k.

How did it go?

Got it moves it further out. Review brings it back tomorrow.