1960 AHSME Problems/Problem 39

Revision as of 11:33, 15 May 2018 by Rockmanex3 (talk | contribs) (Solution to 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 $\boxed{\textbf{(E)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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