Problem on alpha and beta on algebra?

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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)...Show more

(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....Show more