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

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+ | ==Solution 2== | ||

+ | Plug <math>x=1</math> to get <math>a=-1</math>, so <math>x^2-5x+4=0</math>, or <math>(x-4)(x-1)=0</math>, meaning the other solution is <math>x=\boxed{4}</math> | ||

+ | <math>\linebreak</math> | ||

+ | ~Geometry285 | ||

## Revision as of 20:04, 11 July 2021

## Contents

## 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

## 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.