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Worked solution

Real coefficients and a known complex root

Consider the equation 2z3+pz2+qz−65=02z^3+pz^2+qz-65=0.

(i) If pp and qq are real numbers, what are the possibilities for real and complex roots for this equation?

It is given that z=3+2iz=3+2i is a root of the equation 2z3+pz2+qz−65=02z^3+pz^2+qz-65=0, where pp and qq are real numbers.

(ii) Find the values of pp and qq.

[2013/NJC/I/11(a)(modified)]

CJC · Discussion · Question 5

Answer

(i) Three real roots, or one real root and a non-real conjugate pair. (ii) p=−17p=-17, q=56q=56.

(i)

There are three real roots, or one real root and two non-real complex conjugate roots, counting multiplicity. Non-real roots of a real-coefficient polynomial occur in conjugate pairs.

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(ii)

Substitute z=3+2iz=3+2i: 2(3+2i)3+p(3+2i)2+q(3+2i)−65=02(3+2i)^3+p(3+2i)^2+q(3+2i)-65=0. The source evaluates the powers using a graphical calculator: 2(3+2i)3=−18+92i2(3+2i)^3=-18+92i and (3+2i)2=5+12i(3+2i)^2=5+12i. Hence (−18+5p+3q−65)+(92+12p+2q)i=0(-18+5p+3q-65)+(92+12p+2q)i=0. Equate parts: 5p+3q=835p+3q=83 and 6p+q=−466p+q=-46. Solving gives p=−17p=-17, q=56q=56.

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Common mistakes

Count roots with multiplicity; “complex” in the printed contrast means non-real.

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