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

Answer

A, C, D

Explanation

`alpha,beta=(-b+-Delta)/(2a)`
`alpha=(-b-Delta)/(2a),beta=(-b+Delta)/(2a)`
or `alpha=(-b+Delta)/(2a),beta=(-b-Delta)/(2a)`
`implies2aalpha+Delta=-b&2alphabeta-Delta=-b`
`implies2aalpha+Delta=-b&2abeta-Delta-b`
`implies2aalpha+Delta=-b+2Delta`
Sum of roots `=-2b`
sum of roots `=-2b`
product of root `=b^(2)`
product of root `=b^(2)-4`.`(b^(2)-4ac)`
Hence equation is `x^(2)+2bx+b^(2)=0`
`=-3b^(2)+16ac`
Hence equation is `x^(2)+2bx-3b^(2)+16ac=0`