Problem on alpha and beta on algebra?

D=36-4c

x1,2 = (-6 +/- sqr(D))/2 = -3 +/- sqr(36-4c)/2

x2-x1 = (-3+sqr(36-4c)/2) - (-3-sqr(36-4c)/2) = sqr(36-4c) = 2n

36-4c = 4*n^2

9-4c = n^2

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-3-sqr(36-4c)/2 <0, -3 + sqr(36-4c)/2>0

-3-sqr(9-c)<0, -3+sqr(9-c)>0

1st condition: 9-c>0 -> c<9

-3+sqr(9-c)>0 -> sqr(9-c)>3 -> 9-c>9 -> c<0

Therefore 9-4c>9 -> n^2 >3 -> n>sqr(3), n<-sqr(3)

(i)

Let us assume the roots of the given quadratic equation to be l and m.

Now, For a given quadratic equation ax^2+bx+c ,

Sum of roots = -b/a

Product of roots = c/a.

Difference between roots = ( b^2 - 4ac ) / (a^2)

Here, l+m = -6 ; l*m = c ; l-m = 36-4c = 2n(as given in question)

This is a system of linear equations in two variables.On solving, we get

l = n-3 and m = -(n+3).

So, l*m = c (as seen above)

=>

(n-3) * [-(n+3)] = c.

=>

9-n^2 = c.

Or, n^2 = 9-c.

(ii)

Now,

l-m = sqrt[(l+m)^2 - 4lm]

= 36-4c = 2n.

=>

n = 18-2c.

For real roots,We have

36-4c >= 0

=> c <= 9.

=> 2c <= 18.

=> -2c >= -18

=> 18-2c >= 0.

=> n >= 0.

For l > 0 and m < 0,

n-3 > 0 and -(n+3) < 0.

=>

n > 3.

For l < 0 and m > 0,

n-3 < 0 and -(n+3) > 0.

=>

n < -3.

=>

n<-3&n>3.