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If ratio of the roots of the quadratic equation `3m^2x^2+m(m-4)x+2=0` is `lamda` such that `lamda+1/lamda=1` then least value of `m` is (A) `-2-2sqrt3` (B) `-2+2sqrt3` (C) `4+3sqrt2` (D) `4-3sqrt2`

A

`-2+sqrt2`

B

`4-2sqrt3`

C

`4-3sqrt2`

D

`2-sqrt3`

Text Solution

Verified by Experts

The correct Answer is:
C
Let the given quadrartic equation in x,
`3m^(2)x^(2+m(m-4)x+2=0,m ne0` have roots `alpha and beta,` then `alpha=beta=-(m (m-4))/(3m^(2))and alphabeta=(2)/(3m^(2))`
Also, let `alpha/beta=lamda`
Then, `lamda+1/lamda=1impliesalpha/beta+beta/alpha=1" "("given")`
`impliesalpha^(2)+beta^(2)=alphabeta`
`implies (alpha+beta)^(2)=3alpha beta`
`implies(m^(2)(m-4)^(2))/(9m^(4))=3(2)/(3m^(2))`
`implies(m-4)^(2)=10" "[because m ne0]`
`impliesm-4= +-3sqrt2`
`impliesm=4+-3sqrt2`
The least value of `m=4-3sqrt2`
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