SOLUTION: Three consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1, 3, and 5. Find three oth

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Three consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1, 3, and 5. Find three oth      Log On


   



Question 344449: Three consecutive odd integers are such that the square of the third integer is 15 greater than the sum of the squares of the first two. One solution is 1, 3, and 5. Find three other consecutive odd integers that also satisfy the given conditions. What are the integers?
Found 2 solutions by vleith, Fombitz:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+%28x%2B2%29%5E2+=+%28x%2B4%29%5E2-15
x%5E2+%2B+x%5E2+%2B+4x+%2B+4+=+x%5E2+%2B+8x+%2B+16+-15
2x%5E2+%2B+4x+%2B+4+=+x%5E2+%2B+8x+%2B1
x%5E2+-4x+%2B+3=0
%28x-3%29%28x-1%29=0
x = 3 or x = 1
so 1,3,5
and 3,5,7
Now think about what happens on you square numbers.
So -5-3,-1
and -7,-5,-3
are also solutions

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let N,N+2, and N+4 be the integers.
.
.
%28N%2B4%29%5E2=N%5E2%2B%28N%2B2%29%5E2%2B15
N%5E2%2B8N%2B16=N%5E2%2BN%5E2%2B4N%2B4%2B15
N%5E2%2B8N%2B16=2N%5E2%2B4N%2B19
N%5E2-4N%2B3=0
%28N-3%29%28N-1%29=0
Two solutions:
N-3=0
N=3
The other solution is already known, N=1.
The three other integers are 3, 5, and 7.
.
.
.
Verifying,
7%5E2=3%5E2%2B5%5E2%2B15
49=9%2B25%2B15
49=49
True, good solution.