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Difference between revisions of "1953 AHSME Problems/Problem 20"

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==Solution==
 
==Solution==
  
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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>
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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>. Multiplying that by <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==

Latest revision as of 16:40, 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 $x^2+\frac{1}{x^2}+2$. Multiplying that by $x^2$, 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
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All AHSME Problems and Solutions


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