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Then Denominator$=2x+1$

Fraction $=2x+1x $

Reciprocal of the fraction $=x2x+1 $

It is given that the sum fo the fraction and its reciprocal is $22116 $

$∴2x+1x +x2x+1 =22116 $

$⇒x(2x+1)x_{2}+(2x+1)_{2} =2158 ⇒2x_{2}+x5x_{2}+4x+1 =2158 $

$⇒21(5x_{2}+4x+1)=58(2x_{2}+x)⇒105x_{2}+84x+21=116x_{2}+58x$

$⇒(11x+7)(x−3)=0⇒x=3,−117 $

Fraction $=2x+1x $

Reciprocal of the fraction $=x2x+1 $

It is given that the sum fo the fraction and its reciprocal is $22116 $

$∴2x+1x +x2x+1 =22116 $

$⇒x(2x+1)x_{2}+(2x+1)_{2} =2158 ⇒2x_{2}+x5x_{2}+4x+1 =2158 $

$⇒21(5x_{2}+4x+1)=58(2x_{2}+x)⇒105x_{2}+84x+21=116x_{2}+58x$

$⇒11x_{2}−26x−21=0⇒11x_{2}−33x+7x−21=0⇒11x(x−3)+7(x−3)=0$

$⇒(11x+7)(x−3)=0⇒x=3,−117 $

$∵$ $x$ can not be a negative

$⇒x=3$

Hence the fraction $=2x+1x =73 $

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