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Solve the equation $x^{2}+x+\frac{1}{\sqrt{2}}=0$.

Let $t$ be a real number such that $t^2 = at + b$ for some positive integers $a$ and $b$. Then,for any choice of positive integers $a$ and $b$,$t^3$ is never equal to:

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) $TRUE$?
$(A) (2, \frac{7}{2}, 6) \in S$
$(B) \text{If } (3, b, \frac{1}{12}) \in S, \text{ then } |2b| < 1$
$(C) \text{For any given } (a, b, c) \in S, \text{ the system of linear equations } ax + by = 1, bx + cy = -1 \text{ has a unique solution.}$
$(D) \text{For any given } (a, b, c) \in S, \text{ the system of linear equations } (a+1)x + by = 0, bx + (c+1)y = 0 \text{ has a unique solution.}$

The condition for the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ to be equal is

Let $f(x) = (1 + b^2)x^2 + 2bx + 1$ and $m(b)$ be the minimum value of $f(x)$ for a given $b$. As $b$ varies,the range of $m(b)$ is

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