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Difference between revisions of "1993 AHSME Problems/Problem 20"

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== Solution ==
 
== Solution ==
  
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Let <math>r_1</math> and <math>r_2</math> denote the roots of the polynomial.  Then <math>r_1 + r_2 = 3i</math> is pure imaginary, so <math>r_1</math> and <math>r_2</math> have offsetting real parts.  Write <math>r_1 = a + bi</math> and <math>r_2 = -a + ci</math>.   
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Let <math>r_1</math> and <math>r_2</math> denote the roots of the polynomial.  Then <math>r_1 + r_2 = \dfrac{3i}{10}</math> is pure imaginary, so <math>r_1</math> and <math>r_2</math> have offsetting real parts.  Write <math>r_1 = a + bi</math> and <math>r_2 = -a + ci</math>.   
  
 
Now <math>-k = r_1 r_2 = -a^2 -bc + a(c-b)i</math>.  In the case that <math>k</math> is real, then <math>a(c-b)=0</math> so either <math>a=0</math> or that <math>b=c</math>.  In the first case, the roots are pure imaginary and in the second case we have <math>k = a^2+b^2</math>, a positive number.
 
Now <math>-k = r_1 r_2 = -a^2 -bc + a(c-b)i</math>.  In the case that <math>k</math> is real, then <math>a(c-b)=0</math> so either <math>a=0</math> or that <math>b=c</math>.  In the first case, the roots are pure imaginary and in the second case we have <math>k = a^2+b^2</math>, a positive number.

Latest revision as of 13:02, 1 August 2025

Problem

Consider the equation $10z^2-3iz-k=0$, where $z$ is a complex variable and $i^2=-1$. Which of the following statements is true?

$\text{(A) For all positive real numbers k, both roots are pure imaginary} \quad\\ \text{(B) For all negative real numbers k, both roots are pure imaginary} \quad\\ \text{(C) For all pure imaginary numbers k, both roots are real and rational} \quad\\ \text{(D) For all pure imaginary numbers k, both roots are real and irrational} \quad\\ \text{(E) For all complex numbers k, neither root is real}$

Solution

Let $r_1$ and $r_2$ denote the roots of the polynomial. Then $r_1 + r_2 = \dfrac{3i}{10}$ is pure imaginary, so $r_1$ and $r_2$ have offsetting real parts. Write $r_1 = a + bi$ and $r_2 = -a + ci$.

Now $-k = r_1 r_2 = -a^2 -bc + a(c-b)i$. In the case that $k$ is real, then $a(c-b)=0$ so either $a=0$ or that $b=c$. In the first case, the roots are pure imaginary and in the second case we have $k = a^2+b^2$, a positive number.

We can therefore conclude that if $k$ is real and negative, it must be the first case and the roots are pure imaginary.

It's possible to rule out the other cases by reasoning through the cases, but this is enough to show that $\fbox{B}$ is true.

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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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