Difference between revisions of "1982 AHSME Problems/Problem 17"
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+ | ==Problem== | ||
+ | How many real numbers <math>x</math> satisfy the equation <math>3^{2x+2}-3^{x+3}-3^{x}+3=0</math>? | ||
+ | |||
+ | <math>\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3 \qquad \text {(E)} 4</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
Let <math>a = 3^x</math>. Then the preceding equation can be expressed as the quadratic, <cmath>9a^2-28a+3 = 0</cmath> Solving the quadratic yields the roots <math>3</math> and <math>1/9</math>. Setting these equal to <math>3^x</math>, we can immediately see that there are <math>\boxed{2}</math> real values of <math>x</math> that satisfy the equation. | Let <math>a = 3^x</math>. Then the preceding equation can be expressed as the quadratic, <cmath>9a^2-28a+3 = 0</cmath> Solving the quadratic yields the roots <math>3</math> and <math>1/9</math>. Setting these equal to <math>3^x</math>, we can immediately see that there are <math>\boxed{2}</math> real values of <math>x</math> that satisfy the equation. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1982|num-b=16|num-a=18}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 23:36, 29 June 2025
Problem
How many real numbers satisfy the equation
?
Solution
Let . Then the preceding equation can be expressed as the quadratic,
Solving the quadratic yields the roots
and
. Setting these equal to
, we can immediately see that there are
real values of
that satisfy the equation.
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.