SOLUTION: "Word Problems Involving Products and Factoring" Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus 8 is the sum of their

Click here to see ALL problems on Quadratic Equations

Question 251986: "Word Problems Involving Products and Factoring"
Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus 8 is the sum of their squares. can you solve this pls=( tyty so much

Found 2 solutions by Theo, palanisamy:
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
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.

Answer by palanisamy(496) (Show Source):
You can put this solution on YOUR website!
Answer: Let the two consecutive positive integers be n and n+1
Then, (n+n+1)(n+1-n)+8 = n^2 +(n+1)^2
2n+1+8 = n^2 + n^2 + 2n+1
8 = 2n^2
4 = n^2
2 = n
So the required numbers are 2,3