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

Let `Delta^(2)` be the discriminant and `alpha, beta` be the roots of

1.	Let `Delta^(2)` be the discriminant and `alpha, beta` be the roots of the equation `ax^(2) + bx + c = 0`. Then, `2a alpha + Delta and 2 a beta - Delta` can be the roots of the equationA. `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» Correct Answer - A 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`.

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