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> | ||
| + | == Solution 2== | ||
| + | By Descartes' Rule of Signs, there are no sign changes (all coefficients of terms are positive), so there are no positive real roots, or <math>\fbox{B}</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 15:54, 23 April 2020
Contents
Problem 20
The equation 
has:
Solution
By the rational root theorem, 
are possible rational roots. Because 
for 
, so there are no positive roots. We try 
, so 
, so there are no positive real roots; 
Solution 2
By Descartes' Rule of Signs, there are no sign changes (all coefficients of terms are positive), so there are no positive real roots, or .
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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