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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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