1957 AHSME Problems/Problem 45

Problem

If two real numbers $x$ and $y$ satisfy the equation $\frac{x}{y} = x - y$ , then:

$\textbf{(A)}\ {x \ge 4}\text{ or }{x \le 0}\qquad \\ \textbf{(B)}\ {y}\text{ can equal }{1}\qquad \\ \textbf{(C)}\ \text{both }{x}\text{ and }{y}\text{ must be irrational}\qquad\\ \textbf{(D)}\ {x}\text{ and }{y}\text{ cannot both be integers}\qquad\\ \textbf{(E)}\ \text{both }{x}\text{ and }{y}\text{ must be rational}$

Solution

From the initial equation $\tfrac x y = x-y$ , we can solve for $y$ in terms of $x$ as follows: \begin{align*}\ \frac x y &= x-y \\ x &= xy-y^2 \\ y^2-xy+x &= 0 \\ y = \frac{x \pm \sqrt{x^2-4x}}2 \end{align*} Because $y$ is real, the discriminant of the above expression for $y$ must be $\geq 0$ , so $x^2-4x=x(x-4) \geq 0$ . This is true when $x \geq 4$ and $x \leq 0$ , so we choose answer $\boxed{\textbf{(A)}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 44	Followed by Problem 46
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1957 AHSME Problems/Problem 45

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