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 $r$? $\textbf{(A) }0\hspace{20.2em}\textbf{(B) }4$ $\textbf{(C) }108\hspace{19.3em}\textbf{(D) }\text{It could be 0 or 4.}$ $\textbf{(E) }\text{It could be 0 or 108.}\hspace{12em}\textbf{(F) }18$ $\textbf{(G) }-4\hspace{19em}\textbf{(H) } -108$ $\textbf{(I) }\text{It could be 0 or }-4.\hspace{12em}\textbf{(J) }\text{It could be 0 or }-108.$ $\textbf{(K) }\text{It could be 4 or }-4.\hspace{11.5em}\textbf{(L) }\text{There is no such value of }r.$ $\textbf{(M) }1\hspace{20em}\textbf{(N) }-2$ $\textbf{(O) }\text{It could be }-2\text{ or }-4.\hspace{10.3em}\textbf{(P) }\text{It could be 0 or }-2.$ $\textbf{(Q) }\text{It could be 2007 or a yippy dog.}\hspace{6.6em}\textbf{(R) }2007$

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of all possible values of $a$? $\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18$

Let $r$, $s$, $t$ be the roots of the polynomial $x^3+2x^2+x-7$. Then \[ \left(1+\frac{1}{(r+2)^2}\right)\left(1+\frac{1}{(s+2)^2}\right)\left(1+\frac{1}{(t+2)^2}\right)=\frac{m}{n} \] for relatively prime positive integers $m$ and $n$. Compute $100m+n$. [i]Proposed by Justin Stevens[/i]

Equiangular hexagon $ ABCDEF$ has side lengths $ AB \equal{} CD \equal{} EF \equal{} 1$ and $ BC \equal{} DE \equal{} FA \equal{} r$. The area of $ \triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $ r$? $ \textbf{(A)}\ \frac {4\sqrt {3}}{3} \qquad \textbf{(B)}\ \frac {10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac {17}{4} \qquad \textbf{(E)}\ 6$

Let $a,b$ be two positive integers, such that $ab\neq 1$. Find all the integer values that $f(a,b)$ can take, where \[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]

There are 6 distinct values of $x$ strictly between $0$ and $\frac{\pi}{2}$ that satisfy the equation \[ \tan(15 x) = 15 \tan(x) . \] Call these 6 values $r_1$, $r_2$, $r_3$, $r_4$, $r_5$, and $r_6$. What is the value of the sum \[ \frac{1}{\tan^2 r_1} + \frac{1}{\tan^2 r_2} + \frac{1}{\tan^2 r_3} + \frac{1}{\tan^2 r_4} + \frac{1}{\tan^2 r_5} + \frac{1}{\tan^2 r_6} \, ? \]

Let $ d$ and $ e$ denote the solutions of $ 2x^2\plus{}3x\minus{}5\equal{}0$. What is the value of $ (d\minus{}1)(e\minus{}1)$? $ \textbf{(A)}\ \minus{}\frac{5}{2} \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Monic quadratic polynomials $ P(x)$ and $ Q(x)$ have the property that $ P(Q(x))$ has zeroes at $ x\equal{}\minus{}23,\minus{}21,\minus{}17, \text{and} \minus{}15$, and $ Q(P(x))$ has zeroes at $ x\equal{}\minus{}59, \minus{}57, \minus{}51, \text{and} \minus{}49$. What is the sum of the minimum values of $ P(x)$ and $ Q(x)$? $ \textbf{(A)}\ \text{\minus{}100} \qquad \textbf{(B)}\ \text{\minus{}82} \qquad \textbf{(C)}\ \text{\minus{}73} \qquad \textbf{(D)}\ \text{\minus{}64} \qquad \textbf{(E)}\ 0$

Prove that if $a$ and $b$ are positive integers and $ab>1$, then \[\left\lfloor\frac{(a-b)^2-1}{ab}\right\rfloor=\left\lfloor\frac{(a-b)^2-1}{ab-1}\right\rfloor.\]Here $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. [i]Calvin Deng.[/i]

If $m$ and $n$ are the roots of $x^2+mx+n=0$, $m\ne0$, $n\ne0$, then the sum of the roots is: $\textbf{(A)}\ -\dfrac{1}{2} \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \text{Undetermined} $

Let $a$ and $b$ be the solutions to $x^2-7x+17=0$. Compute $a^4+b^4$.

Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^4-18s^2-8s$.

$a,b,c$ are distinct real roots of $x^3-3x+1=0$. $a^8+b^8+c^8$ is $ \textbf{(A)}\ 156 \qquad \textbf{(B)}\ 171 \qquad \textbf{(C)}\ 180 \qquad \textbf{(D)}\ 186 \qquad \textbf{(E)}\ 201$

Find the sum of the $2007$ roots of \[(x-1)^{2007}+2(x-2)^{2006}+3(x-3)^{2005}+\cdots+2006(x-2006)^2+2007(x-2007).\]

Find all positive reals $x,y,z $ such that \[2x-2y+\dfrac1z = \dfrac1{2014},\hspace{0.5em} 2y-2z +\dfrac1x = \dfrac1{2014},\hspace{0.5em}\text{and}\hspace{0.5em} 2z-2x+ \dfrac1y = \dfrac1{2014}.\]