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

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) (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) (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 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:

divide by two to both sides

and

Check 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

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