1960 AHSME Problems/Problem 39

Problem

To satisfy the equation $\frac{a+b}{a}=\frac{b}{a+b}$ , $a$ and $b$ must be:

$\textbf{(A)}\ \text{both rational}\qquad\textbf{(B)}\ \text{both real but not rational}\qquad\textbf{(C)}\ \text{both not real}\qquad$ $\textbf{(D)}\ \text{one real, one not real}\qquad\textbf{(E)}\ \text{one real, one not real or both not real}$

Solution

First, note that $a \neq 0$ and $a \neq -b$ . Cross multiply both sides to get $a^2 + 2ab + b^2 = ab$ Subtract both sides by $ab$ to get $a^2 + ab + b^2 = 0$ From the quadratic formula, $a = \frac{-b \pm \sqrt{b^2 - 4b^2}}{2}$ $a = \frac{-b \pm \sqrt{-3b^2}}{2}$ If $b$ is real, then $\sqrt{-3b^2}$ is imaginary because $-3b^2$ is negative. If $b$ is not real, where $b = m+ni$ and $n \neq 0$ , then $\sqrt{-3b^2}$ evaluates to $\sqrt{-3m^2 - 6mni + 3n^2}$ . As long as $m \neq 0$ , the expression can also be imaginary because a real number squared will be a real number. From these two points, the answer is $\boxed{\textbf{(E)}}$ .

Video Solution

https://youtu.be/ZdM2ou5Gsuw?t=312

~MathProblemSolvingSkills.com

1960 AHSC (Problems • Answer Key • Resources)
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1960 AHSME Problems/Problem 39

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