How many roots does the equation $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$ have?

Let $\alpha, \beta$ be two roots of the quadratic equation $x^2+ax-b=0, b \neq 0$. If the straight line $x \cos \theta + y \sin \theta = c$ touches the curve $(\frac{x}{\alpha})^n + (\frac{y}{\beta})^n = 2$ at the point $(\alpha, \beta)$,then $(\frac{a}{b})^2 + \frac{2}{b} =$

If the roots of the equation $(a^2 + b^2)t^2 - 2(ac + bd)t + (c^2 + d^2) = 0$ are equal,then:

If $a, b, c \in Q$,then the roots of the equation $(b + c - 2a)x^2 + (c + a - 2b)x + (a + b - 2c) = 0$ are

If $(x + 1)$ is a factor of ${x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$,then $p = $

If $f(x)$ is a quadratic expression such that $f(1) + f(2) = 0$,and $-1$ is a root of $f(x) = 0$,then the other root of $f(x) = 0$ is