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.

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) (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:
Simplify and solve for n.

Factor out 25.
So...
Factor.
Apply the zero product principle.
and/or
So, you actually get two multipliers. Let's check both of them:
n = 4
Simplify.

This checks so the three consecutive multiple of 5 are:
20, 25, and 30
Now let's try n = -2
Simplify.

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?

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