SOLUTION: The square of the smaller of two consecutive odd integers exceeds their sum by 97. Find the integers

Question 1058624: The square of the smaller of two consecutive odd integers exceeds their sum by 97. Find the integers
Answer by solve_for_x(190) (Show Source): You can put this solution on YOUR website!
Let x represent the smaller of the two numbers.

The next odd integer is x + 2.

Since the square of the smaller number is 97 more than their sum, you
can write:

x^2 = (x + x + 2) + 97

Rearranging this equation gives:

x^2 - 2x - 99 = 0

This can be factored as:

(x + 9)(x - 11) = 0

which gives the following values:

x = -9, x = 11

Solutions:
1. The numbers are -9 and -7
2. The numbers are 11 and 13
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