Quadratic Equation Solution10





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

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