Difference between revisions of "1953 AHSME Problems/Problem 20"
(Created page with "==Problem 20== If <math>y=x+\frac{1}{x}</math>, then <math>x^4+x^3-4x^2+x+1=0</math> becomes: <math>\textbf{(A)}\ x^2(y^2+y-2)=0 \qquad \textbf{(B)}\ x^2(y^2+y-3)=0\ \tex...") |
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\textbf{(C)}\ x^2(y^2+y-4)=0 \qquad | \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} </math> | \textbf{(D)}\ x^2(y^2+y-6)=0\ \textbf{(E)}\ \text{none of these} </math> | ||
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+ | ==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> | ||
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+ | ==See Also== | ||
+ | {{AHSME 50p box|year=1953|num-b=19|num-a=21}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 23:49, 25 January 2020
Problem 20
If , then becomes:
Solution
We multiply each of the answers to get: , where is either or . Looking at the first term, we have to square , or , doing so, we get the equation . Plugging that into , we get . Multiplying by , we get the expression . Adding these two equations together, we get . To get the term , which was in the original equation, must be , giving an answer of
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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