InterviewSolution
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InterviewSolution
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If one root of `Ax^(3)+Bx^(2)+Cx+D=0,Dne0` is the arithmetic mean of the other two roots, then the relation `2B^(2)+lambdaABC+muA^(2)D=0` holds good. Then, the value of `2lambda+mu` is |
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Answer» Given equation, `Ax^(3)+Bx^(2)+Cx+D=0" " ".......(i)"` where, `Ane 0` Let roots are `alpha, beta, gamma` then `beta=(alpha+gamma)/(2)" " "…….(ii)"` Given relation, `2B^(3)+lambdaABC+muA^(2)D=0" " "....(iii)"` From Eq. (i), `alpha+beta+gamma=-(beta)/(A)` `implies 3beta=- (B)/(A)" " [" from Eq. (ii) "]` `implies beta=- (B)/(3A)` Now, `beta` satisfy Eq. (i), so `A((-B)/(3A))^(3)+B((-B)/(3A))^(2)+C((-B)/(3A))+D=0` `implies (-B^(3))/(27A^(2))+(B^(3))/(9A^(2))-(BC)/(3A)+D=0` `implies (2)/(27)(B^(3))/(A^(2))-(BC)/(3A)+D=0` `implies 2B^(3)-9ABC+27AD^(2)=0` Compare with Eq. (iii), we get `lambda=-9,mu=27` `2lambda +mu=-18+27=9`. |
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