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Worked solution

Evaluate a complex quadratic discriminant exactly

Find the roots of iz2−(5+i)z+2−6i=0iz^2-(5+i)z+2-6i=0, giving answers in Cartesian form a+bia+bi where a,b∈Ra,b\in\mathbb R. marks

[2022/ACJC/II/3(i)]

CJC · Review problems · Question 10

Answer

z=1−3iz=1-3i or z=−2iz=-2i.

Simplify the discriminant

With A=iA=i, B=−(5+i)B=-(5+i), C=2−6iC=2-6i, the discriminant is B2−4AC=(5+i)2−4i(2−6i)=(24+10i)−(24+8i)=2iB^2-4AC=(5+i)^2-4i(2-6i)=(24+10i)-(24+8i)=2i. Since (1+i)2=2i(1+i)^2=2i, its two square roots are ±(1+i)\pm(1+i).

How did this part go?

Use both signs in the quadratic formula

z=5+i±(1+i)2iz=\dfrac{5+i\pm(1+i)}{2i}. The plus sign gives (6+2i)/(2i)=1−3i(6+2i)/(2i)=1-3i; the minus sign gives 4/(2i)=−2i4/(2i)=-2i. Thus the roots are 1−3i1-3i and −2i-2i.

How did this part go?

Common mistakes

Keep the entire discriminant under the radical, and use both signs. Dividing by i multiplies by −i.

Solution notes

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

The source’s middle formula contains √(2i), not 2i; the radical is lost in extraction.

Read the hint