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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.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` |
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Answer» Correct Answer - A::C::D `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` |
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