SOLUTION: Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers.

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Question 169539: Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers.
Found 2 solutions by Mathtut, Edwin McCravy:
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
call the integers a,a+2,a+4



so the integers are 63,65,67
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Three consecutive odd integers are such that the square of the third is 264 more than the square of the second. Find the three integers. 

First odd integer  = 
Second odd integer = 
Third odd integer  = 

the square of the third = 

the square of the second = 

>>...the square of the third is 264 more than the square of the second...<<

Replace "the square of the third" by 
 
>>... is 264 more than the square of the second...<<

Replace "the square of the second" by .

>>... is 264 more than ...<<

Replace the word "is" by :

>>...  264 more than ...<<

Make  264 more than it is by adding 264 to it on the right:

That is, replace "264 more than " by 

>>...  ...<<

So the equation is



Can you solve that equation?  If not post again asking how.

Solution to that equation: x=63

So:

First integer  = 
Second integer = 
Third integer  = 

Answer:  63, 65, 67

Edwin
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