The roots of the equation $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$ for $a \in R$ are always:

If the roots of the equation $x^2 - 8x + a^2 - 6a = 0$ are real,then what is the range of $a$?

For which value of $a$ is the expression $x^2 - ax + 1 - 2a^2$ always positive for all real values of $x$?

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:

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ has

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...