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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Revision as of 19:11, 30 April 2018
Problem 20
If , then becomes:
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