If $a < 0$ then the inequality $a{x^2} - 2x + 4 > 0$ has the solution represented by
$\frac{{1 + \sqrt {1 - 4a} }}{a} > x > \frac{{1 - \sqrt {1 - 4a} }}{a}$
$x < \frac{{1 - \sqrt {1 - 4a} }}{a}$
$x < 2$
$2 > x > \frac{{1 + \sqrt {1 - 4a} }}{a}$
Let $p$ and $q$ be two real numbers such that $p+q=$ 3 and $p^{4}+q^{4}=369$. Then $\left(\frac{1}{p}+\frac{1}{q}\right)^{-2}$ is equal to
If the roots of the equation $8{x^3} - 14{x^2} + 7x - 1 = 0$ are in $G.P.$, then the roots are
If the sum of two of the roots of ${x^3} + p{x^2} + qx + r = 0$ is zero, then $pq =$
Let $\alpha, \beta$ be two roots of the equation $x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0$. Then $\alpha^{8}+\beta^{8}$ is equal to:
If $\log _{(3 x-1)}(x-2)=\log _{\left(9 x^2-6 x+1\right)}\left(2 x^2-10 x-2\right)$, then $x$ equals