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2021 JMPSC Invitationals Problems/Problem 1

Revision as of 14:21, 11 July 2021 by Grisham (talk | contribs) (→‎Solution)

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

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