Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"
(Created page with "==Problem== The equation <math>ax^2 + 5x = 4,</math> where <math>a</math> is some constant, has <math>x = 1</math> as a solution. What is the other solution? ==Solution== asdf") |
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| − | + | Since <math>x=1</math> must be a solution, <math>a+5=4</math> must be true. Therefore, <math>a = -1</math>. We plug this back in to the original quadratic to get <math>5x-x^2=4</math>. We can solve this quadratic to get <math>1,4</math>. We are asked to find the 2nd solution so our answer is <math>\boxed{4}</math> | |
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| + | ~Grisham | ||
Revision as of 15:21, 11 July 2021
Problem
The equation 
where 
is some constant, has 
as a solution. What is the other solution?
Solution
Since 
must be a solution, 
must be true. Therefore, 
. We plug this back in to the original quadratic to get 
. We can solve this quadratic to get 
. We are asked to find the 2nd solution so our answer is 
~Grisham
