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== | ||
+ | Note that there are no sign changes (all coefficients of terms are positive), so by Descartes' rule of signs, there are no positive real roots <math>\Rightarrow \fbox{B}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1954|num-b=19|num-a=21}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 15:55, 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
Note that there are no sign changes (all coefficients of terms are positive), so by Descartes' rule of signs, there are no positive real roots .
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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