Difference between revisions of "1991 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | <math>\fbox{E}</math> Let <math>a = 2^x-4, b = 4^x-2</math>, so the equation becomes <math>a^3+b^3=(a+b)^3 = a^3+3a^2b+3ab^2+b^3 \implies 3a^2b+3ab^2=0 \implies ab(a+b)=0</math>. Hence <math>a=0 \implies 2^x=4 \implies x=2</math>, or <math>b=0 \implies 4^x=2 \implies x=\frac{1}{2}</math>, or <math>a+b=0 \implies 4^x+2^x-6=0 \implies (2^x)^2+2^x-6=0 \implies (2^x+3)(2^x-2)=0</math>, but <math>2^x >0</math> for all <math>x</math>, so we cannot have <math>2^x = -3</math>; however, <math>2^x=2</math> works, giving <math>x=1.</math> Thus we have <math>3</math> solutions: <math>2, 1/2, 1</math>, whose sum is <math>\frac{7}{2}.</math> |
== See also == | == See also == | ||
Revision as of 16:42, 23 February 2018
Problem
The sum of all real 
such that 
is
(A) 3/2 (B) 2 (C) 5/2 (D) 3 (E) 7/2
Solution
Let 
, so the equation becomes 
. Hence 
, or 
, or 
, but 
for all , so we cannot have 
; however, 
works, giving 
Thus we have 
solutions: 
, whose sum is 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
