# 1953 AHSME Problems/Problem 20

## 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 byProblem 19 Followed byProblem 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

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