Difference between revisions of "1953 AHSME Problems/Problem 20"
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==Solution== | ==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} | + | 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> |
==See Also== | ==See Also== |
Revision as of 17:39, 27 June 2022
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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All AHSME Problems and Solutions |
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