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If lambda be the ratio of the roots of t...

If `lambda` be the ratio of the roots of the quadratic equation in x, `3m^2 x^2+m(m-4)x+2=0`, then the least value of m for which `lambda+(1)/(lambda)=1`, is `k-3 sqrt(2)`. The value of k is __________.

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The correct Answer is:
4

`(alpha)/(beta)=lambda`
Sum of roots `=beta lambda +beta=(-m(m-4))/(3m^2)=-((m-4)/(3m)` …..(1)
Product of roots `=beta lambda xxbeta=(2)/(3m^2)` …….(2)
From (1) and (2) `(beta^2(lambda+1)^2)/(beta^2lambda)=((m-4)^2)/(6)`
`((lambda+1)^2)/(lambda)=((m-4)^2)/(6)p ^^ q`
`(lambda^2+2lambda+1)/(lambda)=((m-4)^2)/(6)`
`lambda+(1)/(lambda)=((m-4)^2)/(6)-2`
`lambda+(1)/(lambda)=1 rArr ((m-4)^2)/(6)-2=1`
`(m-4)^2=18`
`m-4=+-3sqrt(2)`
`m=4+-3sqrt(2)`.
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