Difference between revisions of "1954 AHSME Problems/Problem 20"
(→Solution) |
(→Solution 2) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 8: | Line 8: | ||
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== | == 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== | ==See Also== | ||
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 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
