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.
{{{(5
Algebra ->
Sequences-and-series
-> 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.
{{{(5
Log On
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
-----------
25N + 25N + 25 - 125 = 25N + 100
25N - 100 = 100
25N = 200
N = 8
(