Find two consecutive positive odd numbers such that the square of their sum exeeds the sum of their squares by 126

Answers: 9 and 7

Step-by-step explanation:

Givens

The first odd number is n - 1 where n is even but taking one away makes it odd.

The second odd number is n+1 some condition as above except you are adding 1.

Equation

Sum of the odd number is (n - 1 + n + 1)

Their squares summed = (n + 1)^2 + (n - 1)^2

(n + 1 + n - 1)^2 -  (n +1)^2 + (n - 1)^2 = 126

Solution

(2n)^2 - (n^2 + 2n + 1 + n^2 - 2n + 1) = 126     Collect like terms. Expand.

4n^2 -  ( 2n^2 + 2) = 126               Remove the brackets

4n^2 - 2n^2 - 2 = 126                    Collect like terms on the left.

2n^2 - 2 = 126                               Add 2

2n^2 = 126 + 2                              Combine

2n^2 = 128                                    Divide by 2

n^2 = 64                                       Take the square root

sqrt(n^2) = sqrt(64)

n = 8

Answer

n + 1 = 9

n - 1 = 7

Check

Remark

What a really interesting problem. Thanks for posting.

constants mean the stopped points or the points on a graph that nothing is changing, hope this

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what is your question that you ask

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