AS 91577 · 2024 paper · a factor, a power in polar form, a surd equation, a locus gradient, and a line a quotient never meets
Question 1(c) Merit
Practises Polynomial equations with complex roots
Five separate parts: a factor of a cubic, a power of a number in polar form, an equation with two square roots, the gradient of a straight-line locus, and a proof that a quotient never lies on y = x.
Solve the equation 4 − √kx = √(kx + 4) for x, in terms of k, where k is a positive real constant.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 The right side is one root on its own. Square both sides. How many terms does the left side give?
Three: (4 − √kx)2 = 16 − 8√kx + kx.
16 − 8√kx + kx = kx + 4 -
2 The kx terms cancel. Get the root on its own.
12 = 8√kx, so √kx = 3/2. -
3 Square again and solve for x.
kx = 9/4, so x = 9/4k. -
4 Squaring can let in a false answer. Check it in the original.
√kx = 3/2. Left: 4 − 3/2 = 5/2. Right: √(9/4 + 4) = √(25/4) = 5/2 ✓
What the marker wanted
- Achieved Both sides squared correctly, middle term included: 16 − 8√kx + kx = kx + 4.
- Merit The value of x, 9/4k or an equivalent.
In our words, from NZQA’s 2024 marking schedule.
✕Common mistake
NZQA’s 2024 report, in its general comments on the paper, lists solving an equation with surds in it, and multiplying surds correctly, among the things Not Achieved scripts could not do. Here the trap is the square of 4 − √kx: it has a middle term.How did it go?
Got it moves it further out. Review brings it back tomorrow.