SOLUTION: 1. 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. Thank you very much for the answer !

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Question 484238: 1. 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.
Thank you very much for the answer ! :)
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
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

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
1. 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.
Thank you very much for the answer ! :)

Let the smaller # be S
Then the larger # is S + 1

The sum of the #s is S + S + 1, or 2S + 1
The difference of the #s is (S + 1) – S or 1 (this is true as they are consecutive #s)

Product of the sum and difference of the #s = (2S + 1)1, or 2S + 1

We can now say that: 2S+%2B+1+%2B+8+=+S%5E2+%2B+%28S+%2B+1%29%5E2
2S+%2B+9+=+S%5E2+%2B+S%5E2+%2B+2S+%2B+1
0+=+2S%5E2+%2B+2S+-+2S+%2B+1+-+9
2S%5E2+-+8+=+0
2%28S%5E2+-+4%29+=+2%280%29
S%5E2+-+4+=+0

(S + 2)(S – 2) = 0

S = - 2 (ignore as answer has to be a positive number) or S = 2

Since S, or smaller number is 2, then the numbers are highlight_green%282_and_+3%29

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Check
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Sum of 2 and 3: 5 (2 + 3)

Difference of 2 and 3: 1 (3 - 2)

Product of sum and difference: 5 (5 * 1)
Product of the sum and difference of the numbers plus 8 is the sum of their squares is as follows:

5 + 8 = 2%5E2+%2B+3%5E2
13 = 4 + 9
13 = 13 (TRUE)

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