Consider the equation ; x⁴ + ax³ - 6x² + ax + 1

find parameter of a for which equation has two distict positive roots ?

 

The answer is a is (-infinity,2).

Consider only positive side of x-axis for analysis, that is x>0.
At a=2, you get f(x)=x4+2x3-6x2+2x+1,
Differentiating with respect to x: f’(x)=4x3+6x2-12x+2.
When a=2, the minimum value of f(x)=0 at x=1. You can find this by putting f’(x)=0 for a=2.

When a>2, {f(x) when a>2} > {f(x) when a=2} , as the coefficients for a are positive. Hence, by increasing a you are increasing the value for f(x) for any x. Hence minimum value will be more than 0. Hence complex roots on positive side.

When a0 and f(infinity)>0 , you will get two values at which f(x) cuts +ve x -axis. 

Hence a

ATB.