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Solve `6x^(3)-11x^(2)+6x-1=0`, roots of the equatioin are in HP. |
Answer» Put `x=1/y` in the given equation then `6/(y^(3))-11/(y^(2))+6/y-1=0` `impliesy^(3)-6y^(2)+11y-6=0`……..i Now, roots of Eq. (i) are in AP. Let the roots be `alpha-beta, alpha, alpha +beta` The, sum of roots `=alpha-beta+alpha+beta=6` `implies3 alpha=6` `:.alpha=2` Product of roots `=(alpha-beta).alpha.(alpha+beta)=6` `implies(2-beta)2(2+beta)=6implies4-beta^(2)=3` `:.beta=+-1` `:.` Roots of Eqs (i) are 1,2,3 or 3,2,1. Hence roots of the given equation are `1,1/2,1/3` or `1/3, 1/2, 1`. |
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