AS 91577 · 2023 paper · a remainder, an unknown from a quotient, a surd equation, comparing parts, and a pair of conditions
Question 3(c) Merit
Practises Polynomial equations with complex roots
Five separate parts: a remainder, an unknown found from a quotient, an equation with two square roots, two unknowns found from a reciprocal, and a constant pinned down by two conditions at once.
Solve the equation 4√(4x − w) = 5 − 8√x for x, in terms of w.
Work it out
Try it on paper first. Stuck? Take one hint at a time.
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1 The root on the left is on its own already. Square both sides.
16(4x − w) = (5 − 8√x)2 = 25 − 80√x + 64x -
2 Expand the left side. What cancels?
64x − 16w = 25 − 80√x + 64x. The 64x cancels. -
3 Get √x on its own.
80√x = 25 + 16w, so √x = (25 + 16w)/80. -
4 Square to find x.
x = ((25 + 16w)/80)2 = (25 + 16w)2/6400 -
5 Squaring can let in a false answer. When is this a real solution?
√x cannot be negative, and 5 − 8√x is 4 times a root, so it cannot be negative either. Both hold when −25/16 ≤ w ≤ 25/16, and for those w the answer checks exactly. The question asks only for x.
What the marker wanted
- Achieved Both sides squared correctly, middle term included: 64x − 16w = 25 − 80√x + 64x.
- Merit The expression for x, in this form or an equivalent one.
In our words, from NZQA’s 2023 marking schedule.
How did it go?
Got it moves it further out. Review brings it back tomorrow.