lf $α, β$ are the roots of the equation $x^{2} - a x + b = 0$ , then the equation whose roots are $2 α + \frac{1}{β}, 2 β + \frac{1}{α}$ is:

b x^{2} + (2 b + 1) a x + (2 b + 1)^{2} = 0

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b x^{2} - (2 b + 1) a x + (2 b + 1)^{2} = 0

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b x^{2} - (2 b + 1) a x - (2 b + 1)^{2} = 0

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b x^{2} + (2 b + 1) a x - (2 b + 1)^{2} = 0

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Solution

The correct option is D $b x^{2} - (2 b + 1) a x + (2 b + 1)^{2} = 0$
$α + β = \frac{- b}{a} = - (\frac{- a}{1}) = a$ and $α β = b$

Sum of roots $2 α + \frac{1}{β} + 2 β + \frac{1}{α} = 2 (α + β) + \frac{1}{α} + \frac{1}{β}$

$= 2 (+ a) + \frac{a}{b} = 2 a + \frac{a}{b}$

Product of roots $= (2 α + \frac{1}{β}) (2 β + \frac{1}{α})$

$= 4 α β + 4 + \frac{1}{α β} = 2 b + 4 + \frac{1}{b}$

Required equation is
$∴ x^{2} - \frac{(2 b + 1)}{b} a x + (\frac{4 b^{2} + 4 b + 1}{b}) = 0.$

$b x^{2} - (2 b + 1) n + (4 b^{2} + 4 b + 1) = 0$

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lf α, β are the roots of the equation x2−ax+b=0, then the equation whose roots are 2α+1β,2β+1α is:

The correct option is D bx2−(2b+1)ax+(2b+1)2=0α+β=−ba=−(−a1)=a and αβ=bSum of roots 2α+1β+2β+1α=2(α+β)+1α+1β=2(+a)+ab=2a+abProduct of roots =(2α+1β)(2β+1α)=4αβ+4+1αβ=2b+4+1bRequired equation is∴x2−(2b+1)bax+(4b2+4b+1b)=0.bx2−(2b+1)n+(4b2+4b+1)=0

lf $α, β$ are the roots of the equation $x^{2} - a x + b = 0$ , then the equation whose roots are $2 α + \frac{1}{β}, 2 β + \frac{1}{α}$ is: