• # question_answer If $a<0$ then the inequality $a{{x}^{2}}-2x+4>0$ has the solution represented by [AMU 2001] A) $\frac{1+\sqrt{1-4a}}{a}>x>\frac{1-\sqrt{1-4a}}{a}$ B) $x<\frac{1-\sqrt{1-4a}}{a}$ C) x < 2 D) $2>x>\frac{1+\sqrt{1-4a}}{a}$

$a{{x}^{2}}-2x+4>0$ Þ $x=\frac{2\pm \sqrt{4-16a}}{2a}$ Þ $x=\frac{1\pm \sqrt{1-4a}}{a}$  \   $\frac{1-\sqrt{1-4a}}{a}<x<\frac{1+\sqrt{1-4a}}{a}$.