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

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==See also==
 
==See also==
#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]]
#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 
{{JMPSC Notice}}
 
{{JMPSC Notice}}

Revision as of 17:29, 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


See also

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

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