# Difference between revisions of "2007 iTest Problems/Problem 18"

## Problem

Suppose that $x^3+px^2+qx+r$ is a cubic with a double root at $a$ and another root at b, where $a$ and $b$ are real numbers. If $p=-6$ and $q=9$, what is $r$?

## Solutions

### Solution 1

By Vieta's Formulas, $\[2a+b = 6\]$ $\[2ab + a^2 = 9\]$ From the first equation, $b = -2a+6$. Substitute into the second equation to get $\[-4a^2 + 12a + a^2 = 9\]$ $\[-3a^2 + 12a - 9 = 0\]$ $\[a^2 - 4a + 3 = 0\]$ Thus, $a = 3$ or $a = 1$. If $a = 3$, then $b = 0$, so by Vieta's Formulas, $r = 0$. If $a = 1$, then $b = 4$, so by Vieta's Formulas, $r = -4$. The answer is $\boxed{\textbf{(I)}}$.

### Solution 2

The equation of the polynomial with double root $a$ and single root $b$ is $\[(x-a)^2 (x-b)\]$ Expanding the polynomial results in $\[(x^2 - 2ax + a^2)(x-b)\]$ $\[x^3 - (2a+b)x^2 + (2ab+a^2)x - a^2b\]$ Since $p = -6$ and $q = 9$, we can write a system of equations. $\[2a+b = 6\]$ $\[2ab + a^2 = 9\]$ Solving for $a$ results in $a = 3$ or $a = 1$. If $a = 3$, then $b = 0$, so $r = 0$. If $a = 1$, then $b = 4$, so $r = -4$. Thus, the answer is $\boxed{\textbf{(I)}}$.