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

 
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==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>
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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 into 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
 
~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
  
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== Solution 3 ==
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<cmath>ax^2+5x-4=0</cmath>Plugging in <math>1</math>, we get <math>a+5-4=0 \implies a+1=0 \implies a=-1</math>, therefore,
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<cmath>-x^2+5x-4=0 \implies (x-4)(x-1)=0</cmath>Finally, we get the other root is <math>4</math>.
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- kante314 -
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== Solution 4 ==
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We can rearrange the equation to get that <math>ax^2 + 5x - 4 = 0</math>. Then, by Vieta's Formulas, we have <cmath>x = -\frac{4}{a}</cmath> and <cmath>1+x = -\frac{5}{a},</cmath> where <math>x</math> is the second root of the quadratic. Solving for <math>x</math> tells us that the answer is <math>\boxed{4}</math>.
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~Mathdreams
  
 
==See also==
 
==See also==

Latest revision as of 12:42, 12 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 into 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

Solution 3

\[ax^2+5x-4=0\]Plugging in $1$, we get $a+5-4=0 \implies a+1=0 \implies a=-1$, therefore, \[-x^2+5x-4=0 \implies (x-4)(x-1)=0\]Finally, we get the other root is $4$.

- kante314 -

Solution 4

We can rearrange the equation to get that $ax^2 + 5x - 4 = 0$. Then, by Vieta's Formulas, we have \[x = -\frac{4}{a}\] and \[1+x = -\frac{5}{a},\] where $x$ is the second root of the quadratic. Solving for $x$ tells us that the answer is $\boxed{4}$.

~Mathdreams

See also

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

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