If be the ratio of the roots of the quadratic equation in x ,3m^2x^2+m(m-4)x+2=0, then the least value of m for which + 1 =1, is:

Question

Accepted Answer

Correct answer: B. (b) Let roots of the quadratic equation are , Given, = and + 1 =1 ⇒ + =1 ( + )^2-2 αβ αβ=1 … . . (i ) The quadratic equation is, 3m^2x^2+m(m-4)x+2=0 ∴ + = m(4-m)3m^2= 4-m3m and αβ= 23m^2 Put these values in equation (i) ( 4-m3m )^2 23m^2=3 ⇒(m-4)^2=18 ⇒m=4 18 Therefore, least value is 4- 18=4-3 2

$If λ be the ratio of the roots of the quadratic equation in x, 3 m^{2} x^{2} + m (m - 4) x + 2 = 0, then the least value of m for which λ + \frac{1}{λ} = 1, is:$

Solution

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