Consider the equation ${x^2} + \alpha x + \beta = 0$ having roots $\alpha ,\beta $ such that $\alpha \ne \beta $ .Also consider the inequality $\left| {\left| {y - \beta } \right| - \alpha } \right| < \alpha $ ,then
inequality is satisfied by exactly two integral values of $y$
inequality is satisfied by all values of $y \in (-4, 2)$
Roots of the equation are of same sign
${x^2} + \alpha x + \beta > 0\,\forall \,x \in \,\left[ { - 1,0} \right]$
The equation $\sqrt {3 {x^2} + x + 5} = x - 3$ , where $x$ is real, has
Let $\alpha_1, \alpha_2, \ldots, \alpha_7$ be the roots of the equation $x^7+$ $3 x^5-13 x^3-15 x=0$ and $\left|\alpha_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_7\right|$. Then $\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6$ is equal to $..................$.
If $\alpha, \beta $ and $\gamma$ are the roots of the equation $2{x^3} - 3{x^2} + 6x + 1 = 0$, then ${\alpha ^2} + {\beta ^2} + {\gamma ^2}$ is equal to
Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is
Let $p(x)=a_0+a_1 x+\ldots+a_n x^n$ be a non-zero polynomial with integer coefficients. If $p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$, then the smallest possible value of $n$ is