Difference between revisions of "1980 AHSME Problems/Problem 14"
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Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3</math>. | Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3</math>. | ||
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Revision as of 11:48, 5 July 2013
As 
, we can plug that into 
and simplify to get 
. However, we have a restriction on x such that if 
we have an undefined function. We can use this to our advantage. Plugging that value for x into yields 
, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that 
, which is answer choice A.
Alternatively, after simplifying the function to , multiply both sides by 
and divide by 
to yield 
. This can be factored to 
. This means that both 
and either one of 
or 
are equal to 0. 
yields and the other two yield 
. The clear solution is .
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
