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Difference between revisions of "1980 AHSME Problems/Problem 8"
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We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by <math> ab </math>.
 
We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by <math> ab </math>.
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<math> a+b=\frac{ab}{a+b} \\ a^2+2ab+b^2=ab \\ a^2+ab+b^2=0.</math>
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<cmath>a+b=\frac{ab}{a+b} </cmath>
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<cmath>a^2+2ab+b^2=ab </cmath>
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<cmath> a^2+ab+b^2=0.</cmath>
  
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By the quadratic formula, this has no real solutions. <math>\boxed{(A)}</math>
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By the quadratic formula and checking the discriminant (imagining one of the variables to be constant), we see that this has no real solutions. Thus the answer is <math>\boxed{\text{(A)none}}</math>.
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=7|num-a=9}}
 
{{AHSME box|year=1980|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 14:59, 20 March 2018

Problem

How many pairs $(a,b)$ of non-zero real numbers satisfy the equation

\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}\] $\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$ $\text{(E)} \ \text{two pairs for each} ~b \neq 0$

Solution

We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$. \[a+b=\frac{ab}{a+b}\] \[a^2+2ab+b^2=ab\] \[a^2+ab+b^2=0.\]

By the quadratic formula and checking the discriminant (imagining one of the variables to be constant), we see that this has no real solutions. Thus the answer is $\boxed{\text{(A)none}}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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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