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 Accuracy Problems/Problem 4"

(→‎Solution 4)
 
(2 intermediate revisions by 2 users not shown)
Line 10: Line 10:
 
<cmath>(x+8)(x-4) = 0</cmath>
 
<cmath>(x+8)(x-4) = 0</cmath>
  
−
Thus, <math>x = -88</math> or <math>x = 4</math>. Our answer is <math>(-8) \cdot 4=\boxed{-32}</math>
+
Thus, <math>x = -8</math> or <math>x = 4</math>. Our answer is <math>(-8) \cdot 4=\boxed{-32}</math>
  
 
~Bradygho
 
~Bradygho
Line 28: Line 28:
  
 
~tigerzhang
 
~tigerzhang
 +
 +
 +
 +
==See also==
 +
#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
 +
#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 16:23, 11 July 2021

Problem

If $\frac{x+2}{6}$ is its own reciprocal, find the product of all possible values of $x.$

Solution

From the problem, we know that \[\frac{x+2}{6} = \frac{6}{x+2}\] \[(x+2)^2 = 6^2\] \[x^2+ 4x + 4 = 36\] \[x^2 + 4x - 32 = 0\] \[(x+8)(x-4) = 0\]

Thus, $x = -8$ or $x = 4$. Our answer is $(-8) \cdot 4=\boxed{-32}$

~Bradygho

Solution 2

We have $\frac{x+2}{6} = \frac{6}{x+2}$, so $x^2+4x-32=0$. By Vieta's our roots $a$ and $b$ amount to $\frac{-32}{1}=\boxed{-32}$

~Geometry285

Solution 3

$\frac{x+2}{6}=\frac{6}{x+2} \implies x^2+4x-32$ Therefore, the product of the root is $-32$

~kante314

Solution 4

The only numbers that are their own reciprocals are $1$ and $-1$. The equation $\frac{x+2}{6}=1$ has the solution $x=4$, while the equation $\frac{x+2}{6}=-1$ has the solution $x=-8$. The answer is $4 \cdot (-8)=\boxed{-32}$.

~tigerzhang


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy 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_Accuracy_Problems/Problem_4&oldid=158065"