Q19 of 47 Page 1

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

Let the numbers be x, x + 1 and x + 2 (x >0 as x is natural no)


Given:


(x + 1)2 = (x + 2)2 - x2 + 60


x2 + 2x + 1 = x2 + 4x + 4 - x2 + 60


x2 - 2x - 63 = 0


x2 - 9x + 7x - 63 = 0


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


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


x = - 7 or x = 9


Note that x = - 7 is not possible as x is a natural no


So three no's are 9, 10 and 11


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