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Two Natural Number Differ by 3 and Their Product is 504. Find the Numbers.

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Question

Two natural number differ by 3 and their product is 504. Find the numbers. 

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Solution

Let the required numbers be x and (x+3) According to the question: 

`x(x+3)=504` 

⇒`x^2+3x=504` 

⇒`x^2+3x-504=0` 

⇒`x^2+(24-24)x-504=0` 

⇒`x^2+24x-24x-504=0` 

⇒`x(x-24)-21(x+24)=0` 

⇒`(x+24)(x-21)=0` 

⇒`x+24=0  or  x-21=0` 

`⇒x=-24  or  x=21` 

If x = -24, the numbers are `-24 and{(-24+3)=-21}` 

If x =21, the numbers are `21 and {(21+3)=24}`  

Hence, the numbers are `(-24,-21) and (21,24)`

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Chapter 4: Quadratic Equations - Exercises 5

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercises 5 | Q 8

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