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> | ||
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| + | ==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. | ||
Latest revision as of 22:02, 16 June 2021
Problem
How many real numbers 
satisfy the equation 
?
Solution
Let 
. Then the preceding equation can be expressed as the quadratic, ![\[9a^2-28a+3 = 0\]](http://latex.artofproblemsolving.com/4/2/2/422ab93927f22c8a9b3a9f52cbfa13901dc24fe1.png)
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.
