Question

three consecutive natural numbers are such that the square of the middle number exceeds the difference of squares of the Other two
by 60 find the numbers

Answer 1

let first natural number is x
then next consecutives are x+1,x+2
$( {x + 1)}^{2} = ( {x + 2)}^{2} - {x}^{2} + 60 \\ {x}^{2} + 2x + 1 = {x }^{2} + 4x + 4 - {x}^{2} + 60 \\ {x}^{2} + 2x - 4x + 1 - 4 - 60 = 0 \\ {x}^{2} - 2x - 63 = 0 \\ {x}^{2} - 9x + 7x - 63 = 0 \\ x(x - 9) + 7(x - 9) = 0 \\ (x - 9)(x + 7) = 0 \\ x - 9 = 0 \\ x = 9 \: \: \: take \: this \: value \: of \: x \: because \\ \: question \: says \: positive \: number \\ x + 7 = 0 \\ x = - 7 \: discard \: negative \: value \\ so \: numbers \: are \: 9 \\ 10 \\ 11$

Answer 2

Step-by-step explanation:

hope it helps you!

thankyou.