Embibe Experts Solutions for Chapter: Quadratic Equations, Exercise 3: Exercise-3

The quadratic equation $p\left(x\right)=0$ with real coefficients has purely imaginary roots. Then the equation $p\left(p\left(x\right)\right)=0$ has?

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation  $\alpha x^2-x+\alpha =0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $\left|x_1-x_2\right|<1$. Which of the following intervals is (are) a subset(s) of $S$?

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?

If $a\in R$ and the equation $-3{\left(x-\left[x\right]\right)}^2+2\left(x-\left[x\right]\right)+a^2=0$ (where $\left[x\right]$ denotes the greatest integer $\le x$) has no integral solution, then all possible values of $a$ lie in the interval:

Let $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of equation ${\mathrm{px}}^2+\mathrm{qx}+\mathrm{r}=0,\mathrm{p}\ne 0.$ If $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are in $\mathrm{A}.\mathrm{P}.$ and $\frac{1}{\mathrm{\alpha }}+\frac{1}{\mathrm{\beta }}=4,$ then the value of $\left|\mathrm{\alpha }-\mathrm{\beta }\right|$ is 

Let $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of equation ${\mathrm{x}}^2-6\mathrm{x}-2=0.$ If ${\mathrm{a}}_{\mathrm{n}}={\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}},$ for $\mathrm{n}\ge 1,$ then the value of $\frac{{\mathrm{a}}_{10}-2{\mathrm{a}}_8}{2{\mathrm{a}}_9}$ is equal to

The number of all possible positive integral values of $\mathrm{\alpha }$ for which the roots of the quadratic equation, $6{\mathrm{x}}^2-11\mathrm{x}+\mathrm{\alpha }=0$ are rational numbers is

If $\mathrm{\lambda }$ be the ratio of the roots of the quadratic equation in $x, 3m^2{x}^2+m\left(m-4\right)x+2=0,$ then the least value of $m$ for which $\mathrm{\lambda }+\frac{1}{\mathrm{\lambda }}=1,$ is