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

1 Answer

3 votes

Answer:

Answers: 9 and 7

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

  • (9 + 7)^2 = 16^2 = 256
  • 9^2 + 7^2 = 81 + 49 = 130
  • Difference = 256 - 130 = 126

Remark

What a really interesting problem. Thanks for posting.

User Jxmallett
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