SOLUTION: The product of two consecutive odd integers is 10 more than the square of the smaller integer. Find the integers.

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Question 215047: The product of two consecutive odd integers is 10 more than the square of the smaller integer. Find the integers.
Found 2 solutions by stanbon, drj:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The product of two consecutive odd integers is 10 more than the square of the smaller integer. Find the integers.
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1st: 2x-1
2nd: 2x+1
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(2x-1)(2x+1) = (2x-1)^2+10
4x^2-1 = 4x^2 - 4x + 1 + 10
4x = 12
x = 3
1st = 2x-1 = 5
2nd = 2x+1 = 7
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Cheers,
Stan H.

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
The product of two consecutive odd integers is 10 more than the square of the smaller integer. Find the integers.

Step 1. Let n be one odd number and n+2 be the next consecutive odd number

Step 2. Let n*(n+2) be the product of the two odd integers

Step 3. Let n%5E2 be the square of the smaller integer.

Step 4. n*(n+2)=n^2+10 since the product of two consecutive odd integers is 10 more than the square of the smaller integer. Solve yields the following steps:

n%2A%28n%2B2%29=n%5E2%2B10

n%5E2%2B2n=n%5E2%2B10

2n=10 divide by two to both sides

2n=10%2F2

n=5 and n%2B2=7

Check 5%2A7=5%5E2%2B10 which is a true statement

Step 5. The numbers are 5 and 7.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J