Difference between revisions of "1966 AHSME Problems/Problem 33"

Revision as of 12:04, 4 July 2020

Problem

If $ab \ne 0$ and $|a| \ne |b|$ , the number of distinct values of $x$ satisfying the equation

$\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},$

is:

$\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four}$

Solution

$\fbox{D}$

Solution 2

Let $m=\frac{x-a}{b}$ and $n=\frac{x-b}{a}$ then we have $m+n=\frac{1}{m}+\frac{1}{n}$ $m+n=\frac{m+n}{mn}$ Notice that the equation is possible iff $m+n=0$ or $mn=1$ .

If $m+n=0$ then $\frac{x-a}{b}+\frac{x-b}{a}=0$ $\frac{x-a}{b}=\frac{b-x}{a}$ $x=\frac{a^2+b^2}{a+b}$ Which yields $1$ solution for $x$ .

If $mn=0$ then $(\frac{x-a}{b})(\frac{x-b}{a})=1$ $x^2-(a+b)x=0$ Solving the quadratic gets another $2$ solutions for $x$ .

Thus there are $\boxed{\text{(D) three}}$ solutions in total.

~ Nafer

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 29: / Line 29: @@
 Solving the quadratic gets another <math>2</math> solutions for <math>x</math>.
-Thus there are <math>\boxed{D\text{ three}}</math> solutions in total.
+Thus there are <math>\boxed{\text{(D) three}}</math> solutions in total.
 ~ Nafer

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Difference between revisions of "1966 AHSME Problems/Problem 33"

Revision as of 12:04, 4 July 2020

Contents

Problem

Solution

Solution 2

See also