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 "1953 AHSME Problems/Problem 20"

m
m (→‎Solution)
Line 10: Line 10:
 
==Solution==
 
==Solution==
  
−
We multiply each of the answers to get: <math>x^2(y^2)+x^2(y)+nx^2</math>, where <math>n</math> is either <math>-2,-3,-4,</math> or <math>-6</math>. Looking at the first term, we have to square <math>y</math>, or <math>x+\frac{1}{x}</math>, doing so, we get the equation <math>x^2+\frac{1}{x^2}-2</math>. Plugging that into <math>x^2</math>, we get <math>x^4+2x^2+1</math>. Multiplying <math>y</math> by <math>x^2</math>, we get the expression <math>x^3+x</math>. Adding these two equations together, we get <math>x^4+x^3+2x^2+x+1+nx=0</math>. To get the term <math>-4x^2</math>, which was in the original equation, <math>n</math> must be <math>-6</math>, giving an answer of <math>\boxed{D}</math>
+
We multiply each of the answers to get: <math>x^2(y^2)+x^2(y)+nx^2</math>, where <math>n</math> is either <math>-2,-3,-4,</math> or <math>-6</math>. Looking at the first term, we have to square <math>y</math>, or <math>x+\frac{1}{x}</math>, doing so, we get the equation <math>x^2+\frac{1}{x^2}+2</math>. Plugging that into <math>x^2</math>, we get <math>x^4+2x^2+1</math>. Multiplying <math>y</math> by <math>x^2</math>, we get the expression <math>x^3+x</math>. Adding these two equations together, we get <math>x^4+x^3+2x^2+x+1+nx=0</math>. To get the term <math>-4x^2</math>, which was in the original equation, <math>n</math> must be <math>-6</math>, giving an answer of <math>\boxed{D}</math>
  
 
==See Also==
 
==See Also==

Revision as of 16:39, 27 June 2022

Problem 20

If $y=x+\frac{1}{x}$, then $x^4+x^3-4x^2+x+1=0$ becomes:

$\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\\ \textbf{(C)}\ x^2(y^2+y-4)=0 \qquad \textbf{(D)}\ x^2(y^2+y-6)=0\\ \textbf{(E)}\ \text{none of these}$

Solution

We multiply each of the answers to get: $x^2(y^2)+x^2(y)+nx^2$, where $n$ is either $-2,-3,-4,$ or $-6$. Looking at the first term, we have to square $y$, or $x+\frac{1}{x}$, doing so, we get the equation $x^2+\frac{1}{x^2}+2$. Plugging that into $x^2$, we get $x^4+2x^2+1$. Multiplying $y$ by $x^2$, we get the expression $x^3+x$. Adding these two equations together, we get $x^4+x^3+2x^2+x+1+nx=0$. To get the term $-4x^2$, which was in the original equation, $n$ must be $-6$, giving an answer of $\boxed{D}$

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_20&oldid=175397"