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Question 478776: Find the 3 consecutive multiples of 5 so that the square of the third decreased by 5 times the second number is the same as 25 more than twice the product of the first two numbers.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find the 3 consecutive multiples of 5
x, (x+5), (x+10)
:
"the square of the third decreased by 5 times the second number is the same as 25 more than twice the product of the first two numbers."
(x+10)^2 - 5(x+5) = 2x(x+5) + 25
x^2 + 20x + 100 - 5x - 25 = 2x^2 + 10x + 25
combine like terms
x^2 + 15x + 75 = 2x^2 + 10x + 25
Combine on the right
0 = 2x^2 - x^2 + 10x - 15x + 25 - 75
0 = x^2 - 5x - 50
Factors to
(x-10)(x+5) = 0
two solutions
x = 10, 15, 20 are the three numbers
and
x = -5, 0, 5 can be the three numbers also
:
You can check both of these in the original statemet
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