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

(Created page with "==Problem== The equation <math>ax^2 + 5x = 4,</math> where <math>a</math> is some constant, has <math>x = 1</math> as a solution. What is the other solution? ==Solution== asdf") |
(→Solution) |
||

Line 3: | Line 3: | ||

==Solution== | ==Solution== | ||

− | + | Since <math>x=1</math> must be a solution, <math>a+5=4</math> must be true. Therefore, <math>a = -1</math>. We plug this back in to the original quadratic to get <math>5x-x^2=4</math>. We can solve this quadratic to get <math>1,4</math>. We are asked to find the 2nd solution so our answer is <math>\boxed{4}</math> | |

+ | |||

+ | ~Grisham |

## Revision as of 15:21, 11 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