Question

10
14. The difference of two natural numbers is 3 and the difference of their
3/28
reciprocals is -. Find the numbers by quadratic equation​

Answer 1

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.

Answer 2

Answer :

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• The required numbers are 7 and 4 or -4 and -7.

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Explanation :

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Given :

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• The difference of two natural numbers is 3 & the difference of their reciprocals is 3/28.

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To Find :

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• What are the numbers?

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Solution :

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• Let's consider that the first number is n, As their difference is 3. Therefore the second number will be (n - 3).

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$\underline{\clubsuit{\textit{\textbf{\:According\:to\:the\:Question\::}}}}$

$\\ :\implies\:\sf \dfrac{1}{n - 3} - \dfrac{1}{n} = \dfrac{3}{28}$

$\\ \qquad:\implies\:\sf \dfrac{n - \big(n - 3\big)}{n\big(n - 3\big)} = \dfrac{3}{28}$

$\\ :\implies\:\sf \dfrac{\cancel{n} - \cancel{n} + 3}{n^2 - 3n} = \dfrac{3}{28}$

$\\ \qquad:\implies\:\sf \dfrac{\cancel{3}}{n^2 - 3n} = \dfrac{\cancel{3}}{28}$

$\\ :\implies\:\sf \dfrac{1}{n^2 - 3n} = \dfrac{1}{28}$

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By cross multiplying :

$\\ \qquad:\implies\:\sf 28 = n^2 - 3n$

$\\ :\implies\:\sf n^2 - 3n - 28 = 0$

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Splitting the middle term :

$\\ \qquad:\implies\:\sf n^2 - (7 - 4)n - 28 = 0$

$\\ :\implies\:\sf n^2 - 7n - 4n - 28 = 0$

$\\ \qquad:\implies\:\sf n(n - 7) + 4(n - 7) = 0$

$\\ :\implies\:\sf (n - 7) (n + 4) = 0$

$\\ \qquad:\implies\:\sf n - 7 = 0\:\:or\:\: n + 4 = 0$

$\\ :\implies\:\sf n = 0 + 7\:\:or\:\: n = 0 - 4$

$\\ \qquad:\implies\:\underline{\boxed{\bf{\purple{n = 7\:\:or\:\: n = -4}}}}\:\red{\clubsuit}$

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Hence,

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• If n = 7

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⭒First number = n = 7

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⭒Second number = (n - 3) = (7 - 3) = 4

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• If n = -4

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⭒First number = n = -4

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⭒Second number = (n - 3) = (-4 - 3) = -7

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• Therefore, the required numbers are 7 and 4 or -4 and -7.

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