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Worked solution

A quadratic with complex coefficients

Do not use a calculator in answering this question.

Find the roots of the equation w2(1−i)+4w+(10+10i)=0w^2(1-i)+4w+(10+10i)=0, giving your answers in cartesian form a+iba+ib.

[2018/RI/I/7(a)]

CJC · Discussion · Question 6

Answer

w=−3+iw=-3+i or w=1−3iw=1-3i.

Quadratic formula

w=−4±16−4(1−i)(10+10i)2(1−i)w=\dfrac{-4\pm\sqrt{16-4(1-i)(10+10i)}}{2(1-i)}. Since (1−i)(1+i)=2(1-i)(1+i)=2, the discriminant is 16−40(2)=−6416-40(2)=-64. Thus w=−4±8i2(1−i)w=\dfrac{-4\pm8i}{2(1-i)}. Multiply numerator and denominator by 1+i1+i: w=(−1±2i)(1+i)w=(-1\pm2i)(1+i).

How did this part go?

Expand both choices

(−1+2i)(1+i)=−1−i+2i+2i2=−3+i(-1+2i)(1+i)=-1-i+2i+2i^2=-3+i. The other choice is (−1−2i)(1+i)=−1−i−2i−2i2=1−3i(-1-2i)(1+i)=-1-i-2i-2i^2=1-3i.

How did this part go?

Common mistakes

The coefficients are not all real, so the two roots need not be conjugates.

Read the hint