SOLUTION: Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.
This is what I did, but it's wrong.
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Question 330195: Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.
This is what I did, but it's wrong.
25N + 25N + 25 - 125 = 25N + 100
25N - 100 = 100
25N = 200
N = 8
Found 2 solutions by Alan3354, hanah:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.
This is what I did, but it's wrong.
This is ok, but you didn't square any terms.
(n-4)*(n+2) = 0
n = 4
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25N + 25N + 25 - 125 = 25N + 100
25N - 100 = 100
25N = 200
N = 8
Answer by hanah(4) (Show Source): You can put this solution on YOUR website!
N=2
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