AS 91577 · 2025 paper · a quadratic from one root, a surd squared, an equation in √x, a cubic with two unknowns, and simultaneous equations

Question 3(c) Merit

Practises Polynomial equations with complex roots

Five separate parts: a quadratic built from one complex root, the square of a complex number written with surds, an equation in √x, a cubic with two unknown coefficients and one root given, and a pair of simultaneous equations in two complex unknowns.

Solve the equation (1 + 2√x)(3 + 2√x) = 5 + 6√x for x.

The 2025 paper, page 8 ↗ Marking schedule, page 5 ↗

Work it out

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

1 Expand the left side.
(1 + 2√x)(3 + 2√x) = 3 + 2√x + 6√x + 4x = 3 + 8√x + 4x
2 Set it equal to the right side and tidy up.
3 + 8√x + 4x = 5 + 6√x, so 4x + 2√x − 2 = 0, or 2x + √x − 1 = 0.
3 That is a quadratic in √x. Let u = √x.
2u² + u − 1 = 0, so (2u − 1)(u + 1) = 0, and u = 1/2 or u = −1.
4 Can √x be −1?
No: a square root is never negative, so that one goes. √x = 1/2 gives x = 1/4.
If you square instead (√x = 1 − 2x, so x = 1 − 4x + 4x²), you get x = 1/4 or x = 1, and x = 1 is the false one: it gives 15 on the left and 11 on the right.
5 Check x = 1/4 in the original.
√x = 1/2. Left: (1 + 1)(3 + 1) = 8. Right: 5 + 3 = 8 ✓

Answer x = 1/4, and only that.

What the marker wanted

Achieved Expanded and collected, with the root on its own: 2√x = 2 − 4x or an equivalent.
Merit Both possible values found, and the false one, x = 1, thrown out.

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

Common mistake

Keeping both answers. NZQA’s 2025 report names this part: many students did not check whether both values worked, when only one does.

How did it go?

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