# 1992 AHSME Problems/Problem 28

## Problem

Let $i=\sqrt{-1}$. The product of the real parts of the roots of $z^2-z=5-5i$ is $\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$

## Solution

Applying the quadratic formula gives $$z=\frac{-1\pm\sqrt{21-20i}}{2}$$

Let $$\sqrt{21-20i}=a+bi$$ where $a$ and $b$ are real. Squaring both sides and equating real and imaginary parts gives $$a^2-b^2=21$$ $$2ab=-20$$ Substituting $b=-\frac{10}{a}$, letting $a^2=n$ and solving gives $n=25, -4$, from which we see that $a=5$ and $b=-2$.

Replacing the square root in the quadratic formula with $5-2i$ and simplifying gives the two roots $2-i$ and $-3+i$. The product of their real parts is $-6$. The answer is $\fbox{B}$.

## See also

 1992 AHSME (Problems • Answer Key • Resources) Preceded byProblem 27 Followed byProblem 29 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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