2021 JMPSC Invitationals Problems/Problem 1

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

Solution 2

Plug $x=1$ to get $a=-1$, so $x^2-5x+4=0$, or $(x-4)(x-1)=0$, meaning the other solution is $x=\boxed{4}$ $\linebreak$ ~Geometry285

Solution 3

$\[ax^2+5x-4=0\]$Plugging in $1$, we get $a+5-4=0 \implies a+1=0 \implies a=-1$, therefore, $\[-x^2+5x-4=0 \implies (x-4)(x-1)=0\]$Finally, we get the other root is $4$.

- kante314 -