Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"
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+ | ==Solution 2== | ||
+ | Plug <math>x=1</math> to get <math>a=-1</math>, so <math>x^2-5x+4=0</math>, or <math>(x-4)(x-1)=0</math>, meaning the other solution is <math>x=\boxed{4}</math> | ||
+ | <math>\linebreak</math> | ||
+ | ~Geometry285 | ||
Revision as of 21:04, 11 July 2021
Contents
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
Solution 2
Plug to get
, so
, or
, meaning the other solution is
~Geometry285
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.