Difference between revisions of "1980 AHSME Problems/Problem 8"
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<cmath> a^2+ab+b^2=0.</cmath> | <cmath> a^2+ab+b^2=0.</cmath> | ||
| − | By the quadratic formula and checking the discriminant, we see that this has no real solutions. Thus the answer is <math>\boxed{(A) | + | By the quadratic formula and checking the discriminant, we see that this has no real solutions. Thus the answer is <math>\boxed{(A)\text{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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:13, 23 June 2016
Problem
How many pairs 
of non-zero real numbers satisfy the equation
![\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}\]](http://latex.artofproblemsolving.com/7/6/f/76fadc0160a52878b8c0db07e96c1a608f0c3a78.png)
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 
.
![\[a+b=\frac{ab}{a+b}\]](http://latex.artofproblemsolving.com/f/1/4/f1485ce2892c1f1c68422c3b0c5e6c88e87a8618.png)
![\[a^2+2ab+b^2=ab\]](http://latex.artofproblemsolving.com/f/7/9/f792ee5e45d77efbaf8135ec48f3174f2a8b3601.png)
![\[a^2+ab+b^2=0.\]](http://latex.artofproblemsolving.com/6/e/f/6efa3a0cdc55a6f22318821ac901e1eb2c288777.png)
By the quadratic formula and checking the discriminant, we see that this has no real solutions. Thus the answer is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
