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)
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==Solution==
 
==Solution==
asdf
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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>
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~Grisham

Revision as of 15:21, 11 July 2021

Problem

The equation $ax^2 + 5x = 4,$ where $a$ is some constant, has $x = 1$ as a solution. What is the other solution?

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 $\boxed{4}$

~Grisham

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