Question

Let $p,$ $q,$ and $r$ be the distinct roots of the polynomial $x^3 - 22x^2 + 80x - 67.$ There exist real numbers $A,$ $B,$ and $C$ such that $\begin{gathered} \frac{1}{s^3-22s^2+80s-67}\\ =\frac{A}{s-p}+\frac{B}{s-q}\\ \quad+\frac{C}{s-r} \end{gathered}$ for all real numbers $s$ with $s\notin\{p,q,r\}.$ What is $\dfrac1A+\dfrac1B+\dfrac1C?$

$243$

$244$

$245$

$246$

$247$

Answer: B

Concepts:partial fractions Vieta’s Formulas

Difficulty rating: 2390

Solution:

Multiplying the identity by $(s-p)(s-q)(s-r)$ gives $\begin{aligned} 1={}&A(s-q)(s-r)\\ &+B(s-p)(s-r)\\ &+C(s-p)(s-q). \end{aligned}$ Setting $s=p,q,r$ , in turn, yields $\begin{aligned} \frac1A&=(p-q)(p-r),\\ \frac1B&=(q-p)(q-r),\\ \frac1C&=(r-p)(r-q). \end{aligned}$

Adding and expanding gives $\begin{aligned} \frac1A+\frac1B+\frac1C &=p^2+q^2+r^2\\ &\quad-pq-pr-qr. \end{aligned}$ By Vieta's formulas, $p+q+r=22$ and $pq+pr+qr=80$ , so $\begin{aligned} p^2+q^2+r^2&=22^2-2(80)\\ &=324. \end{aligned}$ Therefore the requested value is $324-80=244.$

Thus, B is the correct answer.

2019 AMC 10A Problem 24

Answer: B

Difficulty rating: 2390

Solution:

Problem 24 in Other Years