The value of $a$ for which the quadratic equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ possesses roots with opposite signs,lies in

The equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ is

If $ax^2 + bx + c = 0$ has real and distinct roots,$\alpha$ and $\beta$ where $\beta > \alpha$. Further,if $a > 0, b < 0$,and $c < 0$,then:

If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

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

If $\alpha, \beta, \gamma, \delta, \varepsilon$ are the roots of the equation $x^5+x^4-13x^3-13x^2+36x+36=0$ and $\alpha < \beta < \gamma < \delta < \varepsilon$,then find the value of $\frac{\varepsilon}{\alpha}+\frac{\delta}{\beta}+\frac{1}{\gamma}$.