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Original question & working

A cubic discriminates real from complex roots · source reference

[2020/VJC/Prelim/I/5] · Question pp. 2 · Working pp. 10, 11

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Three distinct real roots require strict inequalities; “only real roots” includes repeated roots.

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Question

(i) One root of x3−3x+c=0x^3-3x+c=0, where cc is real, is 2−3i2-3i. Without using a calculator, find cc and the other roots. [5]

(ii) If cc is a nonzero purely imaginary number, explain whether x3−3x+c=0x^3-3x+c=0 can have real roots. [1]

(iii) Instead let cc be real. By considering y=x3−3xy=x^3-3x, find the range of cc for which x3−3x+c=0x^3-3x+c=0 has only real roots. [2]

[2020/VJC/Prelim/I/5]

Source working

Printed working

(i) Since the coefficients are real, 2+3i2+3i is also a root. Let aa be the third root.

x3−3x+c=(x−(2+3i))(x−(2−3i))(x−a)=((x−2)−3i)((x−2)+3i)(x−a)=((x−2)2+9)(x−a)=(x2−4x+13)(x−a).\begin{aligned}x^3-3x+c&=(x-(2+3i))(x-(2-3i))(x-a)\\&=((x-2)-3i)((x-2)+3i)(x-a)\\&=((x-2)^2+9)(x-a)\\&=(x^2-4x+13)(x-a).\end{aligned}

Comparing the coefficient of xx: −3=4a+13⇒a=−4-3=4a+13\Rightarrow a=-4. Other roots: −4,2+3i-4,2+3i. Constant terms: c=−13a=52c=-13a=52.

Alternative solution:

(2−3i)3−3(2−3i)+c=0,(2-3i)^3-3(2-3i)+c=0, 23+(31)22(−3i)+(32)2(−3i)2+(−3i)3−6+9i+c=0,2^3+\binom31 2^2(-3i)+\binom32 2(-3i)^2+(-3i)^3-6+9i+c=0, 8−36i−54+27i−6+9i+c=0⇒c=52.8-36i-54+27i-6+9i+c=0\Rightarrow c=52.

The conjugate 2+3i2+3i is also a root. Put

x3−3x+52=(x−(2+3i))(x−(2−3i))(x−a).x^3-3x+52=(x-(2+3i))(x-(2-3i))(x-a).

Compare constants: 52=−(2+3i)(2−3i)a=−13a52=-(2+3i)(2-3i)a=-13a, so a=−4a=-4.

(ii) Not possible to have real roots. If λ∈R\lambda\in\mathbb R were a root, λ3−3λ=−c\lambda^3-3\lambda=-c is impossible: LHS is real and RHS is nonzero purely imaginary.

(iii) Consider y=x3−3xy=x^3-3x. For only real roots, it must intersect y=−cy=-c at least twice. Hence −2≤−c≤2-2\leq-c\leq2, equivalently −2≤c≤2-2\leq c\leq2. At the endpoints, a repeated real root accounts for multiplicity.

Teacher note

All-real coefficients imply complex roots occur in conjugate pairs. The margin specifically states that (2−3i)(2-3i), conjugate of (2+3i)(2+3i), is also a root.

Graph labels

Axes x,yx,y, origin 0, extrema (−1,2),(1,−2)(-1,2),(1,-2), curve y=x3−3xy=x^3-3x, horizontal y=−cy=-c and axis ticks −4,−2,2,4,−2,2-4,-2,2,4,-2,2 are retained in native SVG.

Printed answer key

(i) c=52c=52; other roots −4,2+3i-4,2+3i. (iii) −2≤c≤2-2\leq c\leq2.

Native diagrams

The cubic y=x³−3x and horizontal y=−c

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 2; solution pages 10, 11. These are original document page numbers, retained as text metadata.