2021 JMPSC Invitationals Problems/Problem 1


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 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 $\boxed{4}$


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 $\boxed{4}$.


See also

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

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