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Worked solution

Read cubic roots and rotate roots of a complex quadratic

Do not use a graphic calculator in answering this question.

(a) The diagram shows y=f(x)y=f(x) for a cubic polynomial with real coefficients. What can be said about all roots of f(x)=0f(x)=0? marks

(b) The equation 2z2−(7+6i)z+11+ic=02z^2-(7+6i)z+11+ic=0, where cc is a nonzero real number, has a root z=3+4iz=3+4i. Show that c=−2c=-2 and determine the other root in Cartesian form. Hence find the roots of 2w2+(−6+7i)w−11+2i=02w^2+(-6+7i)w-11+2i=0. marks

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Qualitative cubic sketch with an increasing right-hand end and decreasing left-hand end. It crosses the x-axis once on the negative side, has a local maximum on the y-axis above the x-axis, then a local minimum to the right that also stays above the x-axis. Neither turning point touches the axis. The axes have no numerical scales.
Qualitative cubic sketch with an increasing right-hand end and decreasing left-hand end. It crosses the x-axis once on the negative side, has a local maximum on the y-axis above the x-axis, then a local minimum to the right that also stays above the x-axis. Neither turning point touches the axis. The axes have no numerical scales.
Qualitative cubic sketch with an increasing right-hand end and decreasing left-hand end. It crosses the x-axis once on the negative side, has a local maximum on the y-axis above the x-axis, then a local minimum to the right that also stays above the x-axis. Neither turning point touches the axis. The axes have no numerical scales.

[2017/DHS/I/8(a),(b)]

CJC · Review problems · Question 8

Answer

(a) One real root and a non-real conjugate pair. (b) c=−2c=-2; other z-root 12−i\tfrac12-i; w-roots 4−3i,−1−12i4-3i,-1-\tfrac12i.

(a) Interpret the complete sketch

The curve crosses the horizontal axis once, to the left of the vertical axis; both distinct turning points lie strictly above the horizontal axis. Thus there is one real root (negative in this sketch), and the remaining two roots are non-real. Real coefficients force those two roots to form a complex-conjugate pair. The turning-point information rules out confusing this sketch with a triple real root.

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(b) Determine c

Substitute z=3+4iz=3+4i: 2(3+4i)2−(7+6i)(3+4i)+11+ic=02(3+4i)^2-(7+6i)(3+4i)+11+ic=0. The real terms cancel and the imaginary coefficient is 2+c2+c, so c=−2c=-2.

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(b) Other root: source method 1, factorisation

Write 2z2−(7+6i)z+11−2i=[z−(3+4i)](2z−a)2z^2-(7+6i)z+11-2i=[z-(3+4i)](2z-a) with a∈Ca\in\mathbb C. Comparing constants gives a(3+4i)=11−2ia(3+4i)=11-2i, so a=(11−2i)(3−4i)/25=(25−50i)/25=1−2ia=(11-2i)(3-4i)/25=(25-50i)/25=1-2i. Therefore the other root solves 2z−(1−2i)=02z-(1-2i)=0, giving z=1/2−iz=1/2-i.

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(b) Other root: source method 2, sum of roots

Alternatively let the other root be a+bia+bi. The sum is (7+6i)/2=7/2+3i(7+6i)/2=7/2+3i. Thus a+3=7/2a+3=7/2 and b+4=3b+4=3, giving a=1/2a=1/2, b=−1b=-1 and the same root 1/2−i1/2-i.

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(b) Transform to the w-equation

Put z=iwz=iw. Then 2(iw)2−(7+6i)(iw)+11−2i=02(iw)^2-(7+6i)(iw)+11-2i=0 becomes −2w2−(−6+7i)w+11−2i=0-2w^2-(-6+7i)w+11-2i=0. Multiplying by −1-1 gives the required equation. Therefore iw=3+4iiw=3+4i or iw=1/2−iiw=1/2-i, so w=4−3iw=4-3i or w=−1−12iw=-1-\tfrac12i.

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Common mistakes

The conjugate-root theorem applies only to real coefficients. The quadratic in(b) has complex coefficients, so 3−4i is not automatically another root.

Solution notes

Source filename says 2026; printed header says 2025/JC2. Rendered pages determine notation; the original extraction is retained unchanged.

The page image says a∈C; the extraction misreads its complex-number symbol as real. This is an extraction artifact, not a source restriction.

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