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 147776: 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.
Answer by stanbon(75887) (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.
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1st: 5x
2nd: 5x+5
3rd: 5x+10
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EQUATION:
(5x)^2 + (5x+5)^2 = (5x+10)^2 + 125
25x^2 + 25x^2 + 50x + 25 = 25x^2 + 100x + 100 + 125
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4 + 5
x^2 -2x -8 = 0
(x-4)(x+2) = 0
x = 4 or x = -2
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If x = 4:
1st = 20
2nd = 25
3rd = 30
Checking: 20^2 + 25^2 = 30^2+125
400 + 625 = 900 + 125
1025 = 1025
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If x = -2:
1st = -10
2nd = -5
3rd = 0
Checking:
(-10)^2 + (-5)^2 = 0^2 + 125
100 + 25 = 125
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Cheers,
Stan H.
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