Difference between revisions of "1994 AHSME Problems/Problem 25"
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<math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5 </math> | ||
==Solution== | ==Solution== | ||
− | We have two cases to consider: <math>x</math> is positive or <math>x</math> is negative. If <math>x</math> is positive, we have | + | We have two cases to consider: <math>x</math> is positive or <math>x</math> is negative. If <math>x</math> is positive, we have <math>x+y=3</math> and <math>xy+x^3=0</math> |
− | <math>x+ | + | Solving for <math>y</math> in the top equation gives us <math>3-x</math>. Plugging this in gives us: <math>x^3-x^2+3x=0</math>. Since we're told <math>x</math> is not zero, we can divide by <math>x</math>, giving us: <math>x^2-x+3=0</math> |
− | <math> | ||
− | + | The discriminant of this is <math>(-1)^2-4(1)(3)=-11</math>, which means the equation has no real solutions. | |
− | <math>x | + | We conclude that <math>x</math> is negative. In this case <math>-x+y=3</math> and <math>-xy+x^3=0</math>. Negating the top equation gives us <math>x-y=-3</math>. We seek <math>x-y</math>, so the answer is <math>\boxed{(A) -3}</math> |
− | + | -solution by jmania | |
− | + | ==See Also== | |
− | + | {{AHSME box|year=1994|num-b=24|num-a=25}} | |
− | + | {{MAA Notice}} | |
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Revision as of 03:39, 28 May 2021
Problem
If and are non-zero real numbers such that then the integer nearest to is
Solution
We have two cases to consider: is positive or is negative. If is positive, we have and
Solving for in the top equation gives us . Plugging this in gives us: . Since we're told is not zero, we can divide by , giving us:
The discriminant of this is , which means the equation has no real solutions.
We conclude that is negative. In this case and . Negating the top equation gives us . We seek , so the answer is
-solution by jmania
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 25 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.