In this page quadratic equation solution10 we are going to see solution of the word problems of the topic quadratic equation.

Question 17

The denominator of the fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is 73/24. 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 y is 5 more than its numerator x

Then y = (x + 5)

x/(x + 5)

We have to consider the given fraction in the question as x/(x+5)

The sum of the fraction and its reciprocal is 73/24

x/(x+5) + (x + 5)/x = 73/24

x² + (x + 5) 2/x(x + 5) = 73/24

[x² + x² + 2 (x) (5) + 5 2]/[x² + 5 x] = 73/24

[2 x² + 10 x + 25]/[x² + 5 x] = 73/24

24 [2 x² + 10 x + 25] = 73[x² + 5 x]

48 x² + 240 x + 600 = 73 x² + 365 x

73 x² – 48 x²  + 365 x – 240 x – 600 = 0

25 x² + 125 x – 600 = 0

Now we are going to divide the whole equation by 25,so we get

x² + 5 x – 24 = 0

By factoring -24 as we get 8 and -3 as two factors. We can split -24 as 12 (-2) or 6 (-4). But the simplified values of those are not equal to 5.

x² + 8 x – 3x  – 24 = 0

x (x + 8) – 3 (x + 8)  = 0

(x + 8) (x – 3) =0

x + 8 = 0                x – 3 = 0

x = -8                       x = 3

Here x represents the numerator of the fraction. Since the denominator is 5 more than the numerator it must be (5 + 3) = 8

Therefore the required fraction is 3/8

Verification:

The sum of the fraction and its reciprocal is 73/24.

(3/8) + (8/3) = 73/24

(9 + 64)/24 = 73/24

73/24 = 73/24  