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)