2021 JMPSC Invitationals Problems/Problem 1

Revision as of 16:25, 11 July 2021 by Mathdreams (talk | contribs)


The equation $ax^2 + 5x = 4,$ where $a$ is some constant, has $x = 1$ as a solution. What is the other solution?


Since $x=1$ must be a solution, $a+5=4$ must be true. Therefore, $a = -1$. We plug this back in to the original quadratic to get $5x-x^2=4$. We can solve this quadratic to get $1,4$. We are asked to find the 2nd solution so our answer is $\boxed{4}$


See also

  1. Other 2021 JMPSC Invitational Problems
  2. 2021 JMPSC Invitational Answer Key
  3. All JMPSC Problems and Solutions

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