Difference between revisions of "1999 AHSME Problems/Problem 12"

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==Solution==
 
==Solution==
  
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Finding the number of solutions to <math>p(x) = q(x)</math> will find the number of intersections of the two graphs.
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This is also equivalent to the number of roots of <math>p(x) - q(x) = 0</math>.  Since <math>p(x)</math> and <math>q(x)</math> are both fourth degree polynomials with a leading term of <math>x^4</math>, the <math>x^4</math> term will drop out, leaving at most a third degree polynomial (cubic) on the left side.  By the [[Fundamental Theorem of Algebra]], a cubic polynomial can have at most <math>3</math> real solutions, leading to an answer of <math>\boxed{C}</math>. 
  
 
==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=11|num-a=13}}
 
{{AHSME box|year=1999|num-b=11|num-a=13}}

Revision as of 11:08, 8 August 2011

Problem

What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y \equal{} p(x)$ (Error compiling LaTeX. Unknown error_msg) and $y \equal{} q(x)$ (Error compiling LaTeX. Unknown error_msg), each with leading coefficient 1?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Solution

Finding the number of solutions to $p(x) = q(x)$ will find the number of intersections of the two graphs.


This is also equivalent to the number of roots of $p(x) - q(x) = 0$. Since $p(x)$ and $q(x)$ are both fourth degree polynomials with a leading term of $x^4$, the $x^4$ term will drop out, leaving at most a third degree polynomial (cubic) on the left side. By the Fundamental Theorem of Algebra, a cubic polynomial can have at most $3$ real solutions, leading to an answer of $\boxed{C}$.

See Also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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