Question

two natural numbers differ by 3 find the numbers if the sum of their reciprocals is 7 by 10
quadratic equation method​

Answer 1

Answer:

Answer:-

⇢ First natural number = $\large{\boxed{\sf\blue{5}}}$

⇢ Second natural number = $\large{\boxed{\sf\green{2 }}}$

$\rule{100}{2}$

First let's know which numbers are called natural numbers.

Natural numbers:-

• 1,2,3,4,5,.......................etc so,all the positive numbers are called natural numbers.

Given:-

• Difference between two natural numbers = 3
• Sum of their reciprocals = 7/10

To find:-

• What are the numbers?

Let the first natural number be x and second natural number be y.

1st case:-

⇒ x - y = 3

⇒ x = y + 3 .................(eq.1)

2nd case:-

⇒ 1/x + 1/y = 7/10

⇒ (y + x)/xy = 7/10

⇒ 7xy = 10(y + x)

⇒ 7xy = 10y + 10x

• By putting values from (eq.1)

⇒ 7(y + 3)× y = 10y + 10(y + 3)

⇒ 7y² + 21y = 10y + 10y + 30

⇒ 7y² + 21y - 20y - 30 = 0

⇒ 7y² + y - 30 = 0

• By middle term factorization:-

⇒ 7y² - 14y + 15y - 30 = 0

⇒ 7y(y - 2) + 15(y - 2) = 0

⇒ (7y + 15)(y - 2) = 0

⇒ 7y + 15 = 0 or y - 2 = 0

⇒ 7y = -15 or y = 2

⇒ y = -15/7 or y = 2

$\because\sf As, natural \: number \neq negative$

•°• y = 2

$\rule{100}{2}$

• Putting value of y in (eq.1):-

⇢ x = 2 + 3

•°• x = 5

$\rule{100}2$

$\dashrightarrow\sf\underbrace{First \: natural \: number = 5}$

$\dashrightarrow\sf\underbrace{Second \: natural \: number = 2 }$

Answer 2

AnswEr :

$\bf{\purple{\underline{\underline{\bf{Given\::}}}}}$

Two natural numbers differ by 3. If the sum of their reciprocals is 7/10.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The numbers by (quadratic equation method).

$\bf{\orange{\underline{\underline{\bf{Explanation\::}}}}}$

Let the two natural number be R & M.

$\leadsto\tt{R-M=3}\\\\\leadsto\tt{\orange{R=3+M...................(1)}}$

$\bf{\underline{\underline{\bf{According\;to\:the\:question\::}}}}}}$

$\mapsto\sf{\dfrac{1}{R} +\dfrac{1}{M} =\dfrac{7}{10} }\\\\\\\mapsto\sf{\dfrac{1}{3+M}+\dfrac{1}{M} =\dfrac{7}{10} \:\:\:\:\:\big[from(1)\big]}\\\\\\\mapsto\sf{\dfrac{3+M+M}{M(3+M)} =\dfrac{7}{10} }\\\\\\\mapsto\sf{\dfrac{3+2M}{3M+M^{2} } =\dfrac{7}{10} }\\\\\\\mapsto\sf{10(3+2M)=7(3M+M^{2} )}\\\\\\\mapsto\sf{30+20M=21M+7M^{2} }\\\\\\\mapsto\sf{7M^{2} +21M-20M-30=0}\\\\\\\mapsto\sf{\red{7M^{2} +M-30=0}}$

$\dag\bf{\underline{\underline{\bf{Using\:Quadratic\:equation\::}}}}}}$

As given above equation we can compared with ax² + bx + c = 0.

• a = 7
• b = 1
• c = -30

$\mapsto\tt{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }\\\\\\\mapsto\tt{x=\dfrac{-1\pm\sqrt{1^{2}-4*7*(-30) } }{2*7} }\\\\\\\mapsto\tt{x=\dfrac{-1\pm\sqrt{1-28*(-30)} }{14} }\\\\\\\mapsto\tt{x=\dfrac{-1\pm\sqrt{1+840} }{14} }\\\\\\\mapsto\tt{x=\dfrac{-1\pm\sqrt{841} }{14} }\\\\\\\mapsto\tt{x=\dfrac{-1\pm29}{14} }\\\\\\\mapsto\tt{x=\dfrac{-1+29}{14} \:\:\:Or\:\:\:\dfrac{-1-29}{14} }\\\\\\\mapsto\tt{x=\cancel{\dfrac{28}{14}} \:\:\:\:Or\:\:\:\:\dfrac{-30}{14} }\\\\\\\mapsto\tt{\orange{x=2}}$

We know that negative value isn't acceptable.

Thus,

The 1st natural number is R = 3+2 = 5

The 2nd natural number is M = 2.