Difference between revisions of "1980 AHSME Problems/Problem 8"

(Solution)
(Solution)
 
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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{\text{(A)none}}</math>.
+
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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:59, 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 (ProblemsAnswer KeyResources)
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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