Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
 
(One intermediate revision by one other user not shown)
Line 3: Line 3:
  
 
==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>
+
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
Line 12: Line 12:
 
~Geometry285
 
~Geometry285
  
 +
== Solution 3 ==
 +
<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,
 +
<cmath>-x^2+5x-4=0 \implies (x-4)(x-1)=0</cmath>Finally, we get the other root is <math>4</math>.
 +
 +
- kante314 -
 +
 +
== Solution 4 ==
 +
 +
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>.
 +
 +
~Mathdreams
  
 
==See also==
 
==See also==

Latest revision as of 11:42, 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Invitationals_Problems/Problem_1&oldid=158243"