2021 JMPSC Invitationals Problems/Problem 1

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

Problem

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

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 $\boxed{4}$

~Grisham

See also

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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Invalid username
Login to AoPS