SOLUTION: I have tried to work these problems different ways and can not come up with the correct answer. Any help would be great! Thanks! 1) Find three consecutive integers such that t

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Question 189204: I have tried to work these problems different ways and can not come up with the correct answer. Any help would be great! Thanks!
1) Find three consecutive integers such that the sum of the squares of the smaller two is equal to the square of the largest.
2) Find three consecutive odd integers such that the sum of all three is 36 less than the product of the smaller two.

Found 2 solutions by jonvaliente, josmiceli:
Answer by jonvaliente(64) About Me  (Show Source):
You can put this solution on YOUR website!
Question #1:
Let x, x+1, x+2 = 3 consecutive integers
x%5E2%2B%28x%2B1%29%5E2=%28x%2B2%29%5E2
x%5E2%2Bx%5E2%2B2x%2B1=x%5E2%2B4x%2B4
2x%5E2%2B2x%2B1=x%5E2%2B4x%2B4
2x%5E2-x%5E2%2B2x-4x%2B1-4=0
x%5E2-2x-3=0
%28x-3%29%28x%2B1%29=0
x-3=0 x+1=0
x=3 x=-1
The possible answers are:
3,4,5 or -1,0,1
Question #2:
Let x, x+2, x+4 = consecutive odd numbers
x%2B%28x%2B2%29%2B%28x%2B4%29=x%2A%28x%2B2%29-36
3x%2B6=x%5E2%2B2x-36
x%5E2-x-42=0
%28x-7%29%28x%2B6%29=0
x-7=0 x+6=0
x=7 or x=-6
Since we are looking for odd numbers, the correct answer should be
7,9,11

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(1)
Call the 3 integers a, a+%2B+1, and a+%2B+2
a%5E2+%2B+%28a+%2B+1%29%5E2+=+%28a+%2B+2%29%5E2
a%5E2+%2B+a%5E2+%2B+2a+%2B+1+=+a%5E2+%2B+4a+%2B+4
Subtract a%5E2 from both sides
a%5E2+%2B+2a+%2B+1+=+4a+%2B+4
Subtract 4a from both sides
a%5E2+-+2a+%2B+1+=+4
Subtract 4 from both sides
a%5E2+-+2a+-+3+=+0
%28a+-+3%29%28a+%2B+1%29+=+0
a+=+3
a+=+-1 Can't use the negative root
a+=+3
a+%2B+1+=+4
a+%2B+2+=+5
The consecutive numbers are 3,4, and 5
(2)
If the integers are odd, they are every other number:
a, a+%2B+2, a+%2B+4
a+%2B+a+%2B+2+%2B+a+%2B+4+=+a%2A%28a+%2B+2%29+-+36
3a+%2B+6+=+a%5E2+%2B+2a+-+36
Subtract 3a from both sides
6+=+a%5E2+-+a+-+36
Subtract 6 from both sides
a%5E2+-+a+-+42+=+0
%28a+-+7%29%28a+%2B+6%29+=+0
a+=+7
a+=+-6 Can't use negative root
a+=+7
a+%2B+2+=+9
a+%2B+4+=+11
the consecutive odd integers are 7, 9, and 11
check:
a+%2B+a+%2B+2+%2B+a+%2B+4+=+a%2A%28a+%2B+2%29+-+36
7+%2B+9+%2B+11+=+7%2A9+-+36
27+=+63+-+36
27+=+27
OK