In the quadratic equation `ax^2 + bx + c = 0`, if `Delta = b^2-4ac and alpha + beta, alpha^2 + beta^2, alpha^3 + beta^3` are in GP. where `alpha, beta` are the roots of `ax^2 + bx + c =0`, then

Options

A. (1,-1)
B. (1,1)
C. (-1/6 , -7/6)
D. (1/6 , 7/6)

Answer

C

Explanation

Since `ax^(2) + bx + 10 = 0` has imaginary roots .
Therefore ,
f(x) = `ax^(2) + bx + 10 gt 0` or , `lt` 0 for all x `in` R
We have , f(0) = 10 `gt 0`
`therefore f(x) = ax^(2) + bx + 10 gt 0` for all x `in` R
`implies f(5) = 25a + 5b + 10 gt` 0
`implies 5a + b + 2 gt 0`
`implies 5a + b gt - 2 `
`implies ` Minimum value of 5a + b is -2 .
`implies 5alpha + beta = -2 [therefore alpha and beta ` are the values of a and b for which 5a + b is minimum ]
`implies beta = -5 alpha -2 `
Now ,
`alpha (4x + 2y + 3) + beta (x - y -1 ) = 0`
`implies alpha ( 4x + 2y + 3) - (5 alpha + 2) ( x - y - 1) = 0 `
`implies alpha (-x + 7y + 8) - 2 (x - y - 1) = 0`
Clearly , it represents a family of lines passing through the intersection of the lines
-x + 7y + 8 = 0 and x - y - 1 = 0 i.e., the point (-1/6 , -7/6).