Difference between revisions of "1980 AHSME Problems/Problem 8"
Mrdavid445 (talk | contribs) (Created page with "==Problem== How many pairs <math>(a,b)</math> of non-zero real numbers satisfy the equation <cmath> \frac{1}{a} + \frac{1}{b} = \frac{1}{a+b} </cmath> <math>\text{(A)} \ \text...") |
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<math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0</math> | <math>\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0</math> | ||
<math>\text{(E)} \ \text{two pairs for each} ~b \neq 0</math> | <math>\text{(E)} \ \text{two pairs for each} ~b \neq 0</math> | ||
| + | |||
| + | == 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 <math> ab </math>. | ||
| + | <math> a+b=\frac{ab}{a+b} \\ a^2+2ab+b^2=ab \\ a^2+ab+b^2=0.</math> | ||
| + | |||
| + | By the quadratic formula, this has no real solutions. <math>\boxed{(A)}</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1980|num-b=7|num-a=9}} | ||
Revision as of 19:34, 31 March 2013
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 
.
By the quadratic formula, this has no real solutions. 
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 | ||
