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

(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...")
 
(Solution 2)
 
(4 intermediate revisions by 3 users not shown)
Line 6: Line 6:
  
 
== 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, </math>x=-3, x=-2<math>, so there are no positive real roots; </math>\fbox{B}$
+
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

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; $\fbox{B}$

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 $\Rightarrow \fbox{B}$.

See Also

1954 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png