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.

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Question 75320This question is from textbook Alegbra 2 An Incremental Development
: 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 question is from textbook Alegbra 2 An Incremental Development

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Here's one approach:
Let n be the multiplier. The first multiple of 5 is 5n, the next consecutive multiple of 5 is 5n+5, and the third consecutive multiple of 5 is 5n+10.
From the problem description, you can write:
%285n%29%5E2%2B%285n%2B5%29%5E2+=+%285n%2B10%29%5E2%2B125 Simplify and solve for n.
25n%5E2%2B25n%5E2%2B50n%2B25+=+25n%5E2%2B100n%2B100%2B125
25n%5E2-50n-200+=+0 Factor out 25.
25%28n%5E2-2n-8%29+=+0 So...
n%5E2-2n-8+=+0 Factor.
%28n-4%29%28n%2B2%29+=+0 Apply the zero product principle.
n+=+4 and/or n+=+-2
So, you actually get two multipliers. Let's check both of them:
n = 4
%285%2A4%29%5E2%2B%285%2A4%2B5%29%5E2+=+%285%2A4%2B10%29%5E2%2B125 Simplify.
20%5E2%2B25%5E2+=+30%5E2%2B125
400%2B625+=+900%2B125
1025+=+1025 This checks so the three consecutive multiple of 5 are:
20, 25, and 30
Now let's try n = -2
%285%28-2%29%29%5E2%2B%285%28-2%29%2B5%29%5E2+=+%285%28-2%29%2B10%29%5E2%2B125 Simplify.
%28-10%29%5E2%2B%28-5%29%5E2+=+%28-10%2B10%29%5E2%2B125
100%2B25+=+0%2B125
125+=+125 This also checks so could the three consecutive multiples of 5 also be?
-10, -5, 0 Well...not really because 0 is not a multiple of 5, is it?