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

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~Grisham
 
~Grisham
  
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==Solution 2==
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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>
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<math>\linebreak</math>
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~Geometry285
  
  

Revision as of 20:04, 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

Solution 2

Plug $x=1$ to get $a=-1$, so $x^2-5x+4=0$, or $(x-4)(x-1)=0$, meaning the other solution is $x=\boxed{4}$ $\linebreak$ ~Geometry285


See also

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

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