# 2016 AMC 12A Problems/Problem 24

## Problem

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 $b$? $\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

## Solution

### Solution 1 (calculus)

The acceleration must be zero at the $x$-intercept; this intercept must be an inflection point for the minimum $a$ value. Derive $f(x)$ so that the acceleration $f''(x)=0$: $x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).

The function with the minimum $a$: $$f(x)=\left(x-\frac{a}{3}\right)^3$$ $$x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}$$ Since this is equal to the original equation $x^3-ax^2+bx-a$, $$\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}$$ $$b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}$$

The actual function: $f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$ $f(x)=0\rightarrow x=\sqrt{3}$ triple root. "Complete the cube."

### Solution 2

Note that since both $a$ and $b$ are positive, all 3 roots of the polynomial are positive as well.

Let the roots of the polynomial be $r, s, t$. By Vieta's $a=r+s+t$ and $a=rst$.

Since $r, s, t$ are positive we can apply AM-GM to get $\frac{r+s+t}{3} \ge \sqrt{rst} \rightarrow \frac{a}{3} \ge \sqrt{a}$. Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields $\frac{a^2}{27} \ge 1 \rightarrow a \ge 3\sqrt{3}$.

Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$. For this value of $a$, we can use Vieta's to get $b=\boxed{\textbf{(B) }9}$.

### Solution 3

All three roots are identical. Therefore, comparing coefficients, the root of this cubic function is $\sqrt{3}$. Using Vieta's, the coefficient we desire is the sum of the pairwise products of the roots. Because our root is unique, the answer is simply $b=\boxed{\textbf{(B) }9}$. (note that this is only true since for the min value of a, applying AM-GM to the sums and products of roots, equality condition produced min value of a )

## See Also

 2016 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 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 All AMC 12 Problems and Solutions

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