Question

The denominator of a fraction is one more than twice its nominator.If the sum of the fraction and it's reciprocal is 2. 16/21 find the fraction.

Answer 1

Answer:

Hi there !!

Let the numerator be x

Given,

The denominator of a fraction is one more than twice it's numerator,

So, denominator = 2x + 1

The original fraction will be

\frac{x}{2x + 1}

2x+1

x

Also,

the sum of the fraction and it's reciprocal is

2 \frac{16}{21} = \frac{58}{21}2

21

16

=

21

58

Reciprocal of the original fraction

= \frac{2x + 1}{x}=

x

2x+1

Their sum is 58/21

So,

the following balanced equation will be formed

\frac{x}{2x + 1} + \frac{2x + 1}{x} = \frac{58}{21}

2x+1

x

+

x

2x+1

=

21

58

Taking LCM as x(2x + 1),

we have,

\frac{x(x) + (2x + 1)(2x + 1)}{x(2x + 1)} = \frac{58}{21}

x(2x+1)

x(x)+(2x+1)(2x+1)

=

21

58

Multiplying the terms,

we've

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

2x

2

+x

x

2

+4x

2

+4x+1

=

21

58

21(x² + 4x² + 4x + 1) = 58(2x² + x)

21(5x² + 4x + 1) = 58(2x² + x)

105x² + 84x + 21 = 116x² + 58x

0 = 116x² - 105x² + 58x - 84x - 21

0 = 11x² - 26x - 21

Splitting the middle term

and using the rule where ab = 11 × -21 = -231 = -33 × 7 and a + b = -33 + 7 = -26

we have,

0 = 11x² - 33x + 7x - 21

Taking out common factors,

we've,

0 = 11x(x - 3) + 7(x - 3)

0 = (11x + 7)(x - 3)

11x + 7 = 0 x - 3 = 0

11x = -7 x = 0 + 3 = 3

x = -7/11.

Since the denominator of a fraction can't be taken as -7/11 because it can't be negative

we'll take x = 3

Numerator = x = 3

Denominator = 2x + 1 = 3 × 2 + 1 = 6 + 1 = 7

Thus,

the fraction is

\frac{3}{7}

7

3

Answer 2

Appropriate 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