SOLUTION: The Sum of the squares of two consecutive odd integers is 74. Find the integers. I have found the integers, 5 and 7. but i must solve it algebraically. Can you help? thank you!

Question 531612: The Sum of the squares of two consecutive odd integers is 74. Find the integers.
I have found the integers, 5 and 7. but i must solve it algebraically. Can you help?
thank you!:)
Found 3 solutions by richard1234, solver91311, stanbon:
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
Let x and x+2 be the integers (x is odd). Then

Just solve for x. Hint: {5,7} is not the only set of solutions.
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

If is an odd integer then the next odd integer must be .

The square of the first is . The square of the other is

The sum of the squares is

Solve the quadratic. It factors. By the way 5 and 7 is only half right.

John

My calculator said it, I believe it, that settles it

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
The Sum of the squares of two consecutive odd integers is 74. Find the integers.
-----
1st odd: 2x-1
2nd odd: 2x+1
-----
Equation:
(2x-1)^2 + (2x+1)^2 = 74
4x^2-4x+1 + 4x^2+4x+1 = 74
-----
8x^2+2 = 74
8x^2 = 72
x^2 = 9
x = 3 or x = -3
----
If x = 3,
1st: 2x-1 = 5
2nd: 2x+1 = 7
----
If x = -3,
1st: 2x-1 = -7
2nd: 2x+1 = -5
======================
Cheers,
Stan H.
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