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  • Three consecutive natural number are such that the square of the middle number exceeds the difference of the square of the other two by 60. Find the numbers       </

Three consecutive natural number are such that the square of the middle number exceeds the difference of the square of the other two by 60. Find the numbers 

 

 

 

 

 
 
 
 
 

Answers (1)

Let the 3 numbers are x , x+1 , x+2 

According to question 

(x+1)^2 = ( x+2)^2 - x^2 + 60 \\\\ x^2 + 1^2 + 2x = x^2 + 2^2 + 4x - x^2 + 60\\\\x^2 + 2x + 1 = 4 + 4x + 60 \\\\ x^2 + 2x + 1 - 4x - 64 = 0 \\\\ x^2 - 2x -63= 0\\\\ x ^2 - (9-7) x-63 = 0 \\\\ x^2 - 9 x + 7x - 63 = 0 \\\\ x (x-9)+ 7 (x-9) = 0

( x-9) ( x+7) = 0

x = 9    or x = -7  

since -7 is not natural no. , hence x = 9 

The natural numbers are 9,10,11

Posted by

Ravindra Pindel

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