Question

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b?$

$8$

$9$

$10$

$11$

$12$

Answer: B

Concepts:Vieta’s Formulas AM-GM Inequality

Difficulty rating: 2380

Solution:

By Vieta's formulas, the real roots $r,s,t$ satisfy $r+s+t=a,$ $rs+st+tr=b,$ and $rst=a,$ so $r+s+t=rst.$

The positive product means either all three roots are positive or two are negative. In the latter case write the roots as $-u,-v,w$ with $u,v,w\gt0.$ The positive sum forces $w\gt u+v,$ and then $b=uv-w(u+v)\lt0,$ a contradiction. Hence all three roots are positive.

By the AM-GM inequality, $27rst\le(r+s+t)^3=(rst)^3,$ so $a=rst\ge 3\sqrt3,$ with equality if and only if $r=s=t=\sqrt3.$ At this smallest $a,$ $\begin{gathered} b=rs+st+tr\\ =3r^2=3\cdot 3=9. \end{gathered}$

Thus, the correct answer is B.

2016 AMC 12A Problem 24

Answer: B

Difficulty rating: 2380

Solution:

Problem 24 in Other Years