Hi aakriti I have already answered it in your other post but here i am pasting the reply again.
Before answering your problem i would like you ppl to look at roots of quadratic equation in a different way.
I want to understand the nature of roots of equation x^2 +(b/a)x+(c/a)=0
What i would do i will define f(x)=x^2 +(b/a)x+(c/a) and now i want to solve for f(x)=0.
You can appriciate that the form of f(x) will be such whose value is infinity at x= -infinity, will reach a minima in between and will again become infinite at x= +infinity.
Now if min(f(x) > 0 then there is no real solution for above equation
if min(f(x) = 0 then there is one real solution for above equation
if min(f(x) <0 then there are two real solution for above equation
Minima of f(x) occurs at x= -b/2a
and value of f(x) at this point is -b^2/(4a)+(c/a)
and hence i can find the nature of root. (The answer will be same if you do it by discriminant method.)
Now look at Aakriti's problem
f(x)=x^4-4x^3-8x^2+r=0
Putting f'(x)=4x^3-12x^2-16x=0
4x(x^2-3x-4)=0
This is zero at x=-1 (This will be first local minima as you move from -infinity to plus infinity.Call f(-1) as minima 1)
x=0 (This will be a local maxima. Call f(0) maxima)
and x=4 (This will be second local minima. Call f(4) minma 2)
If minima1>0 and minma2>0 Then function is always greater then zero and no real roots
If minma1<0, maxima>0 and minima2<0 tThen there are four distinct root
If one of the minima <0 and other >0 or if both minma and maxima are less than zer than only two real roots (Think of it by drawing graphs of such functions)