Quadratic Equation Solution11





In this page quadratic equation solution11 we are going to see solution of the word problems of the topic quadratic equation.

Question 18

The denominator of the fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58/21, find the fraction.

Solution:

Let “x/y” be the required fraction. Here x is known as numerator and y is known as denominator

Here the denominator is one more than twice the numerator

Then y = 2 x + 1

x/(2 x + 1)

We have to consider the given fraction in the question as x/(2 x + 1)

The sum of the fraction and its reciprocal is 58/21

x/(2 x + 1) + (2 x + 1)/x = 58/21

[x² + (2 x + 1) 2]/[x (2 x + 1)] = 58/21

[x² + 4 x² + 2 (2 x) (1) + 1²]/[2 x² +  x] = 58/21

[5 x² + 4 x + 1]/[2 x² +  x] = 58/21

21 [5 x² + 4 x + 1] = 58[2 x² +  x]

105 x² + 84 x + 21 = 116 x² + 58 x

116 x² - 105 x² + 58 x – 84 x – 21 = 0

11 x² – 26 x – 21 = 0

11 x² + 7 x – 33 x – 21 = 0

x (11 x + 7) – 3 ( 11 x + 7) = 0

(x – 3) (11 x + 7) = 0

x – 3 = 0                           11 x + 7 = 0

x = 3                                     11 x = -7

                                              x = -7/11                

The negative value of x is not possible. So we can take 3 for x.

Here x represents the numerator of the fraction. Since the denominator is one more than twice the numerator

By applying x = 3 in y = 2x + 1

We get y = 2(3) + 1

               y = 7           

Therefore the required fraction is 3/7

Verification:

 The sum of the fraction and its reciprocal is 58/21

(3/7) + (7/3) = 58/21

 (9 + 49)/21 = 58/21

58/21 = 58/21

quadratic equation solution11 quadratic equation solution11