# Difference between revisions of "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 into 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 -

## Solution 4

We can rearrange the equation to get that $ax^2 + 5x - 4 = 0$. Then, by Vieta's Formulas, we have $$x = -\frac{4}{a}$$ and $$1+x = -\frac{5}{a},$$ where $x$ is the second root of the quadratic. Solving for $x$ tells us that the answer is $\boxed{4}$.

~Mathdreams

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