Solve the equation $x^{2}+3x+9=0$.

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2}x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+ax+b=0$. Then the roots of the equation $x^{2}-(a+b-2)x+(a+b+2)=0$ are...

If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$ where $ac \neq 0$, then $P(x) \cdot Q(x) = 0$ has $(a, b, c, d \in \mathbb{R})$:

Let $R^2$ denote $R \times R$. Let $S = \{(a, b, c) : a, b, c \in R \text{ and } ax^2 + 2bxy + cy^2 > 0 \text{ for all } (x, y) \in R^2 - \{(0, 0)\}\}$. Then which of the following statements is (are) $TRUE$?
$(A) (2, \frac{7}{2}, 6) \in S$
$(B) \text{If } (3, b, \frac{1}{12}) \in S, \text{ then } |2b| < 1$
$(C) \text{For any given } (a, b, c) \in S, \text{ the system of linear equations } ax + by = 1, bx + cy = -1 \text{ has a unique solution.}$
$(D) \text{For any given } (a, b, c) \in S, \text{ the system of linear equations } (a+1)x + by = 0, bx + (c+1)y = 0 \text{ has a unique solution.}$

Suppose $a, b, c$ are real numbers,and each of the equations $x^2+2ax+b^2=0$ and $x^2+2bx+c^2=0$ has two distinct real roots. Then,the equation $x^2+2cx+a^2=0$ has

If $AM$ and $GM$ of roots of a quadratic equation are $5$ and $4$,respectively,then the quadratic equation is