SOLUTION: Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus eight is the sum of their squares. THANK YOU!!!!!!

Question 63910This question is from textbook Algebra 1
: Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus eight is the sum of their squares.
THANK YOU!!!!!! This question is from textbook Algebra 1

Found 2 solutions by venugopalramana, joyofmath:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find two consecutive positive numbers
N AND N+1 SAY
such that the product of the sum(N+N+1=2N+1) and difference(N+1-N=1) of the numbers plus eight
(2N+1)1+8
is the sum of their squares[(N^2+(N+1)^2].
2N+1+8=N^2+(N+1)^2=N^2+N^2+1+2N
2N^2-8=0
N^2=8/2=4
N=+2 0R -2
HENCE THE 2 NUMBERS ARE
2 AND 3 .....CHECK...(2+3)(3-2)+8=2^2+3^2.....13=13..OK
OR
-2 AND -1..........CHECK.....(-2-1)(-1+2)+8=(-2)^2+(-1)^2...OR....5=5..OK

THANK YOU!!!!!!

Answer by joyofmath(189) (Show Source): You can put this solution on YOUR website!
Find two consecutive positive numbers such that the product of the sum and difference of the numbers plus eight is the sum of their squares.

Let n = the first of the two numbers.
Then n+1 = the second of the numbers.
The equation to solve is .
This expands to .
Or, .
Or, .
Or, .
Or, .
Or, .
This factors: .
So, n=-3 or n=1.
But, the problem states that the numbers are positive.
So, n=1. The second number is n+1 = 2.
We verify this answer by noting that the sum of the numbers (1+2=3) times the difference of the numbers (1-2=-1) plus 8 = 3(-1)+8 = 5 which is indeed the sum of 1 squared plus 2 squared.
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