α,β are the roots of x2−3x+2=0 then α1+α,β1+β are the roots of

α=1,β=2 α1+α=11+1=12,β1+β=21+2=23 x2−(12+23)x+12.23=0 x2−(3+46)x+13=0 6x2−7x+2=0

α,β are the roots of x2−3x+2=0 then α1+α,β1+β are the roots of

α=1,β=2 α1+α=11+1=12,β1+β=21+2=23 x2−(12+23)x+12.23=0 x2−(3+46)x+13=0 6x2−7x+2=0

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$α, β$ are the roots of $x^{2} - 3 x + 2 = 0$ then $\frac{α}{1 + α}, \frac{β}{1 + β}$ are the roots of

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6 x^{2} + 5 x + 2 = 0

6 x^{2} - 7 x + 2 = 0

6 x^{2} - 5 x + 2 = 0

6 x^{2} - 7 x - 2 = 0

answer is B.

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Detailed Solution

α = 1, β = 2

$\frac{α}{1 + α} = \frac{1}{1 + 1} = \frac{1}{2}, \frac{β}{1 + β} = \frac{2}{1 + 2} = \frac{2}{3}$

$x^{2} - (\frac{1}{2} + \frac{2}{3}) x + \frac{1}{2} . \frac{2}{3} = 0$

$x^{2} - (\frac{3 + 4}{6}) x + \frac{1}{3} = 0$

$6 x^{2} - 7 x + 2 = 0$

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α,β are the roots of x2−3x+2=0 then α1+α,β1+β are the roots of

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$α, β$ are the roots of $x^{2} - 3 x + 2 = 0$ then $\frac{α}{1 + α}, \frac{β}{1 + β}$ are the roots of