Math, asked by harukiduggal7752, 10 months ago

The denominator of the fraction is one more than twice the denominator. The sum of the fraction and its reciprocal is 2 16/21, find the fraction.

Answers

Answered by SUBRATA4322
5

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)x2+(2x+1)2=2158⇒2x2+x5x2+4x+1=2158

⇒21(5x2+4x+1)=58(2x2+x)⇒105x2+84x+21=116x

⇒11x2−26x−21=0⇒11x2−33x+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

The answer is 73 .

Hope this answer helps you.

Answered by mathdude500
3

Appropriate Question :-

The denominator of a fraction is 1 more than twice its numerator. If the sum of fraction and its reciprocal is 2\frac{16}{21}  , find the fraction.

\large\underline{\sf{Solution-}}

Let assume that

Numerator of a fraction be x.

So,

Denominator of a fraction be 1 + 2x

Thus,

\sf \: Fraction =  \dfrac{x}{1 + 2x}  \\  \\

According to statement, sum of fraction and its reciprocal is 2\frac{16}{21}  .

So,

\sf \: \dfrac{x}{1 + 2x} +  \frac{2x + 1}{x} = 2 \frac{16}{21}    \\  \\

\sf \: \dfrac{ {x}^{2}  +  {(2x + 1)}^{2} }{(1 + 2x)x}  =  \frac{58}{21}    \\  \\

\sf \: \dfrac{ {x}^{2}  +   {4x}^{2} + 4x + 1}{x + 2 {x}^{2} }  =  \frac{58}{21}    \\  \\

\sf \: \dfrac{{5x}^{2} + 4x + 1}{x + 2 {x}^{2} }  =  \frac{58}{21}    \\  \\

\sf \: 58( {2x}^{2} + x) = 21( {5x}^{2} + 4x + 1) \\  \\

\sf \:  {116x}^{2} + 58x ={105x}^{2} + 84x + 21\\  \\

\sf \:  {116x}^{2} + 58x - {105x}^{2} -  84x  - 21 = 0\\  \\

\sf \:  {11x}^{2} - 26x  - 21 = 0\\  \\

\sf \:  {11x}^{2} - 33x + 7x  - 21 = 0\\  \\

\sf \: 11x(x - 3) + 7(x - 3) = 0 \\  \\

\sf \: (x - 3)(11x + 7) = 0 \\  \\

\sf \:  \implies \: x = 3 \:  \:  \: or \:  \:  \: x =  -  \frac{7}{11} \:  \{rejected \} \\  \\

Hence,

\sf \: \sf \:  \implies \: Fraction =  \dfrac{x}{1 + 2x}  =  \frac{3}{7}  \\  \\

\rule{190pt}{2pt}

Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

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