Question

What’s the answer for this

Answer 1

Let the three consecutive natural numbers be x,x+1,x+2.

Given that the square of the middle number exceeds the difference of the squares of the other two by 60.

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

x^2 + 1 + 2x = 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 - 9)(x + 7) = 0

x = 9 (or) x = -7.

Since x cannot be -ve, so x = 9.


1st number = 9

2nd number = x + 1 = 10

3rd number = x + 2 = 11.


Therefore the three numbers = 9,10,11.


Hope this helps!

Answer 2

$PLAYING \: WITH \: NUMBERS \: \\ \\let \: the \: numbers \: be \: - \\ \\ x - 1 \: \: \: and\: \: \: \: x \: \: and \: \: x + 1 \\ \\ according \: \: to \: \: the \: \: question \: \\ \\ {x}^{2} = {(x + 1)}^{2} - {(x - 1)}^{2} + 60 \\ \\ {x}^{2} = {x}^{2} + 2x + 1 - ( {x}^{2} - 2x + 1) + 60 \\ \\ {x}^{2} = 4x + 60 \\ \\ {x}^{2} - 4x - 60 = 0 \\ \\ (x - 10)(x + 6) = 0 \\ \\ given \: that \: \: \: \: x \: \: \: \: is \: \: a \: natural \: no. \\ \\ x = 10 \\ \\ \\ hence \: the \: \: numbers \: are \: \: \: - \\ \\ 9 \: \: \: and \: \: \: 10 \: \: \: and \: \: \: \: 11 \: \: \: \: \: \: ans.$