Complex Numbers
Real and Imaginary Parts of Roots of Complex Quadratic
nta_pyq_2025_apr
Grade 11
Question:
If $\alpha+i\beta$ and $\gamma+i\delta$ are the roots of $x^2-(3-2i)x-(2i-2)=0$, where $\alpha,\beta,\gamma,\delta$ are real numbers, then $\alpha\gamma+\beta\delta$ equals
Step-by-Step Solution
Key Concept: Find the roots explicitly by inspection or the quadratic formula, then read off $\alpha,\beta,\gamma,\delta$ and compute the dot product directly.
By Vieta's: sum $= (\alpha+\gamma)+i(\beta+\delta) = 3-2i$ and product $= (\alpha+i\beta)(\gamma+i\delta)=-(2i-2)=2-2i$.
Try $z=1$: $1-(3-2i)\cdot1-(2i-2)=1-3+2i-2i+2=0$ ✓.
So one root is $1=1+0i$, and the other is $(3-2i)-1=2-2i$.
Thus $\alpha=1,\beta=0$ and $\gamma=2,\delta=-2$ (or vice versa).
$\alpha\gamma+\beta\delta = 1\cdot2+0\cdot(-2)=2$.
Correct Answer: 4