Let `Delta^(2)` be the discriminant and `alpha,beta` be the roots of the equation `ax^(2)+bx+c=0` then `2aalpha+Delta` and `2abeta-Delta` can be roots of the equation.

Options

A. `x^(2) + 2bx + b^(2) = 0`
B. `x^(2) - 2bx + b^(2) = 0`
C. `x^(2) + 2bx - 3b^(2) - 16 ac = 0`
D. `x^(2) - 2bx - 3b^(2) + 16 ac = 0`

Answer

A

Explanation

We have, `alpha,beta =(-b +- Delta)/(2a)`
Now, two case arise:
CASE I When `alpha = (-b + Delta)/(2a) and, beta = (-b-Delta)/(2a)`
`rArr" "2a alpha + Delta = -b + 2 Delta and, 2 alpha beta - Delta = -b - 2 Delta`
So, the required equation is `x^(2) - x(-2b) + b^(2) - 4 Delta^(2) = 0`
`rArr" "x^(2) + 2bx + b^(2) - 4(b^(2) - 4ac) = 0`
`rArr" "x^(2) + 2bx - 3b^(2) + 16 ac = 0`
CASE II When `alpha = (-b-Delta)/(2a) and, beta=(-b+Delta)/(2a)`
`rArr" "2a alpha + Delta = -b and 2 a beta - Delta = - b`
So, the required equation is `x^(2) - x(-2b)+b^(2) = 0 or, x^(2) + 2bx + b^(2) = 0`.