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 !
Algebra.Com
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) (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) (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:
(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
---------
Check
---------
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 =
13 = 4 + 9
13 = 13 (TRUE)
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