If the roots of ${x^2} + x + a = 0$exceed $a$, then
$2 < a < 3$
$a > 3$
$ - 3 < a < 3$
$a < - 2$
If $x$ is real and $k = \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}},$ then
If $\alpha , \beta , \gamma$ are roots of equation $x^3 + qx -r = 0$ then the equation, whose roots are
$\left( {\beta \gamma + \frac{1}{\alpha }} \right),\,\left( {\gamma \alpha + \frac{1}{\beta }} \right),\,\left( {\alpha \beta + \frac{1}{\gamma }} \right)$
If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)\,p^2 -2p\, (ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$, then
If $2 + i$ is a root of the equation ${x^3} - 5{x^2} + 9x - 5 = 0$, then the other roots are
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