Question

the denominator of a fraction is one more than twice its numerator if the sum of the fraction and reciprocal is 2 into 16 upon 21 find the fraction​

Answer 1

Question :-

The denominator of a fraction is one 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