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

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==See also==
 
==See also==
#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Invitational Problems]]
#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Invitational 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 17:25, 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 Invitational Problems
  2. 2021 JMPSC Invitational Answer Key
  3. All JMPSC Problems and Solutions

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

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