# Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"

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+ | == Solution 3 == | ||

+ | <cmath>ax^2+5x-4=0</cmath>Plugging in <math>1</math>, we get <math>a+5-4=0 \implies a+1=0 \implies a=-1</math>, therefore, | ||

+ | <cmath>-x^2+5x-4=0 \implies (x-4)(x-1)=0</cmath>Finally, we get the other root is <math>4</math>. | ||

+ | |||

+ | - kante314 - | ||

==See also== | ==See also== |

## Revision as of 10:06, 12 July 2021

## Problem

The equation where is some constant, has as a solution. What is the other solution?

## Solution

Since must be a solution, must be true. Therefore, . We plug this back in to the original quadratic to get . We can solve this quadratic to get . We are asked to find the 2nd solution so our answer is

~Grisham

## Solution 2

Plug to get , so , or , meaning the other solution is ~Geometry285

## Solution 3

Plugging in , we get , therefore, Finally, we get the other root is .

- kante314 -

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.