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1014. The difference of two natural numbers is 3 and the difference of their3/28reciprocals is -. Find the numbers by quadratic equation​

Answer»

Answer:

7 and 4

Step-by-step EXPLANATION:

Let the required numbers are 'x' and '3 - x'  as their sum is x+(3-x) = 3.

 GIVEN, difference of their  reciprocals is 3/28.

⇒ 1/(x - 3) - 1/x = 3/28

⇒ (x - (x-3))/x(x - 3) = 3/28

⇒ 3/x(x - 3) = 3/38

⇒ 1/x(x - 3) = 1/28

⇒ 28 = x(x - 3)

⇒ 0 = x² - 3x -28  

    Here, a = 1,  b = -3,   c = -28

Using quadratic formula,

x = [-(-3) ± √(-3)² - 4(1)(-28) ]/2(1)

  = [3 ± √9 + 112 ] /2

  = [3 ± √121]/2

  = (3 + 11)/2      or   (3 - 11)/2

  = 7     or    -4

Hence the required numbers are:

if x = 7,  x - 3 = 4

if x = -4,  x -3 = -7

  As the required numbers are natural, numbers are 7 and 4.


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