Question

Three consecutive natural no are such that the sq of the middle no exceeds the difference of the sq of the other no by 60 find the no.

Answer 1

Let the nos. be X, X + 1 and X + 2.
Now we form equations according to the data given to us.

${(x + 1)}^{2} = {(x + 2)}^{2} - {x}^{2} + 60 \\ {x}^{2} + 2x + 1 = {x}^{2} + 4x + 4 - {x}^{2} + 60 \\ {x}^{2} + 2x + 1 = 4x + 64 \\ {x}^{2} - 2x - 63 = 0 \\ {x}^{2} - 9x + 7x - 63 = 0 \\ x(x - 9) + 7(x - 9) = 0 \\ (x + 7)(x - 9) = 0$
So we find that X can be (-7) or (+9)

But we are looking for natural numbers so X cannot be negative.

Thus X = 9 and so the numbers are

9, 10 and 11