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The difference of two natural numbers is 3 and the difference of their reciprocals is 3 / 28. Find the numbers.Class 10 Quadratic Equations

Answer»

According to the question,

Let the required TWO numbers are x and ( x - 3 ),


Given, the difference of their RECIPROCALS is 3 / 28.

Reciprocal of x = 1 / x

Reciprocal of x - 3 = 1 / ( x - 3 )


⇒ Difference = 3 / 28

\implies \dfrac{1}{x-3} - \dfrac{1}{x}= \dfrac{3}{28}\\\\\\\implies \dfrac{x-x+3}{x(x-3)}=\dfrac{3}{28}\\\\\\\implies \dfrac{3}{x(x-3)}=\dfrac{3}{28}\\\\\\\implies \dfrac{1}{x^2-3x} =\dfrac{1}{28}\\\\\\\implies 28 = x^2-3x


⇒ x^2 - 3x - 28 = 0

⇒ x^2 - ( 7 - 4 )x - 28 = 0

⇒ x^2 - 7x + 4x - 28 = 0

⇒ x( x - 7 ) + 4( x - 7 ) = 0

⇒ ( x - 7 )( x + 4 ) = 0

⇒ x = 7 or - 4


Given that the required numbers are NATURAL numbers, so they can't be negative.

∴ x = 7


Other NUMBER = x - 3 = 7 - 3 = 4


Therefore, required natural numbers are 7 and 4 .


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