2021 JMPSC Invitationals Problems/Problem 1

Revision as of 10:06, 12 July 2021 by Kante314 (talk | contribs)


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 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 $\boxed{4}$


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 -

See also

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

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