If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign,what is their product?

If the roots of $4x^2 + px + 9 = 0$ are equal,then the absolute value of $p$ is:

For $n > 2$ and $n \in N$,the product of the roots of $(x-n)((x^2-2nx)^2 + (2n^2-5)(x^2-2nx) + (n^4-5n^2+4)) = 0$ is divisible by

Let $-\frac{\pi}{6} < \theta < -\frac{\pi}{12}$. Suppose $\alpha_1$ and $\beta_1$ are the roots of the equation $x^2 - 2x \sec \theta + 1 = 0$ and $\alpha_2$ and $\beta_2$ are the roots of the equation $x^2 + 2x \tan \theta - 1 = 0$. If $\alpha_1 > \beta_1$ and $\alpha_2 > \beta_2$,then $\alpha_1 + \beta_2$ equals

Suppose that $x$ and $y$ are positive numbers with $xy = \frac{1}{9}$,$x(y + 1) = \frac{7}{9}$,and $y(x + 1) = \frac{5}{18}$. The value of $(x + 1)(y + 1)$ is equal to:

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are