# 2007 iTest Problems/Problem 18

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## 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 $r$? $\text{(A) }0\qquad \text{(B) }4\qquad \text{(C) }108\qquad \text{(D) It could be }0 \text{ or } 4\qquad \text{(E) It could be }0 \text{ or } 108$ $\text{(F) }18\qquad \text{(G) }-4\qquad \text{(H) }-108\qquad \text{(I) It could be } 0 \text{ or } -4$ $\text{(J) It could be } 0 \text{ or } {-108} \qquad \text{(K) It could be } 4 \text{ or } {-4}\qquad \text{(L) There is no such value of } r\qquad$ $\text{(M) } 1 \qquad \text{(N) } {-2} \qquad \text{(O) It could be } 4 \text{ or } -4 \qquad \text{(P) It could be } 0 \text{ or } -2 \qquad$ $\text{(Q) It could be } 2007 \text{ or a yippy dog} \qquad \text{(R) } 2007$

## 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 $\boxed{\textbf{(I)}}$.