Question 251986
let:


x = first number
x+1 = second number


product of (sum + diff) + 8 = sum (squares)


sum = x + x + 1 = 2x+1


diff = x + 1 - x = 1


product of sum and diff = (2x+1)*1 = 2x+1


sum of the squares is equal to the first number squared plus the second number squared.


deriving the sum of the squares, we get:


x^2 + (x+1)^2 equals:


x^2 + x^2 + 2x + 1


adding these together and combining like terms, we get:


x^2 + x^2 + 2x + 1 equals:


2x^2 + 2x + 1


since sum(squares) = product of sum and diff + 8, we get:


2x^2 + 2x + 1 = 2x + 1 + 8 which becomes:


2x^2 + 2x + 1 = 2x + 9


subtract 2x from both sides of the equation and subtract 1 from both sides of the equation to get:


2x^2 = 8


divide both sides of the equation by 2 to get:


x^2 = 4


take the square root of both sides of the equation to get:


x = sqrt(4) = 2


answer should be x = 2


substitute in original problem statement equations to see if they are true.


sum of two numbers is 2 + 3 = 5


diff of two numbers is 3 - 2 = 1


product of sum and diff is 5*1 = 5


sum of squares = product of sum and diff + 8 becomes:


2^2 + 3^2 = 5 + 8 which becomes:


4 + 9 = 13


since this statement is true, the value for x is good.


your answer is:


the 2 consecutive numbers that will satisfy the problem statement are 2 and 3.