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 ! :)


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 + 1 + 8 = S^2 + (S + 1)^2}}}

{{{2S + 9 = S^2 + S^2 + 2S + 1}}}

{{{0 = 2S^2 + 2S - 2S + 1 - 9}}}

{{{2S^2 - 8 = 0}}}

{{{2(S^2 - 4) = 2(0)}}}

{{{S^2 - 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(2_and_ 3)}}}


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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 = {{{2^2 + 3^2}}}

13 = 4 + 9

13 = 13 (TRUE)


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