Question 484238
Find two consecutive positive numbers
x, (x+2)
:
such that the product of the sum and difference of the numbers plus 8 is the sum of their squares.
(x+(x+2))*(2) + 8 = x^2 + (x+2)^2
(2x + 2) * 2 + 8 = x^2 + x^2 + 4x + 4
4x + 4 + 8 = 2x^2 + 4x + 4
4x + 12 = 2x^2 + 4x + 4
0 = 2x^2 + 4x - 4x + 4 - 12
2x^2 - 8 = 0
2x^2 = 8
x^2 = 8/2
x^2 = 4
x = 2 and 4 are the consecutive numbers
:
:
see if that works in the statement
"the product of the sum and difference of the numbers plus 8 is the sum of their squares. "
(4+2)*(4-2) + 8 = 2^2 + 4^2
6(2) + 8 = 4 + 16