Question

Solve the following inequality:
$\frac{19-33x}{7x^2-11x+4}>2$

Answer 1

$\frac{19-33x}{7x^2-11x+4}>2$

$\frac{19-33x}{7x^2-11x+4}-2>0$

$\frac{-14x^2-11x+11}{7x^2-11x+4}>0$

$\frac{-14x^2-11x+11}{(x-1)(7x-4)}>0$

$-14x^2-11x+11iszerofor:x=-\frac{11+\sqrt{737}}{28},x=\frac{\sqrt{737}-11}{28}$

$x-1iszerofor:x=1$

$7x-4iszerofor:x=\frac{4}{7}$

$-\frac{11+\sqrt{737}}{28}<x<\frac{4}{7}or\frac{\sqrt{737}-11}{28}<x<1$