Difference between revisions of "1954 AHSME Problems/Problem 20"
Katzrockso (talk | contribs) (Created page with "== Problem 20== The equation <math>x^3+6x^2+11x+6=0</math> has: <math> \textbf{(A)}\ \text{no negative real roots}\qquad\textbf{(B)}\ \text{no positive real roots}\qquad\te...") |
Katzrockso (talk | contribs) (→Solution) |
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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, | + | 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> |
Revision as of 12:19, 6 June 2016
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; 
