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Difference between revisions of "1954 AHSME Problems/Problem 20"

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== Solution ==
 
== Solution ==
 
By the rational root theorem, <math>1, -1, 2, -2, 3, -3, 6, -6</math> are possible rational roots. Because <math>x^3+6x^2+11x+6>0</math> for <math>x>0</math>, so there are no positive roots. We try <math>-1, -2, -3, -6</math>, so <math>x=-1, x=-3, x=-2</math>, so there are no positive real roots; <math>\fbox{B}</math>
 
By the rational root theorem, <math>1, -1, 2, -2, 3, -3, 6, -6</math> are possible rational roots. Because <math>x^3+6x^2+11x+6>0</math> for <math>x>0</math>, so there are no positive roots. We try <math>-1, -2, -3, -6</math>, so <math>x=-1, x=-3, x=-2</math>, so there are no positive real roots; <math>\fbox{B}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 00:29, 28 February 2020

Problem 20

The equation $x^3+6x^2+11x+6=0$ has:

$\textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\textbf{(C)}\ \text{no real roots}\\ \textbf{(D)}\ \text{1 positive and 2 negative roots}\qquad\textbf{(E)}\ \text{2 positive and 1 negative root}$

Solution

By the rational root theorem, $1, -1, 2, -2, 3, -3, 6, -6$ are possible rational roots. Because $x^3+6x^2+11x+6>0$ for $x>0$, so there are no positive roots. We try $-1, -2, -3, -6$, so $x=-1, x=-3, x=-2$, so there are no positive real roots; $\fbox{B}$

See Also

1954 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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