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Question 820501: Three brothers have ages that are consecutive even integers. The product of the first and third boy's ages is 20 more than twice the second boys age, find the age of each of the three boys.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Three brothers have ages that are consecutive even integers.
x, (x+2), (x+4) are the ages of the 3 brothers
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The product of the first and third boy's ages is 20 more than twice the second boys age,
x(x+4) = 2(x+2) + 20
x^2 + 4x = 2x + 4 + 20
x^2 + 4x = 2x + 24
Combine as a quadratic equation on the left
x^2 + 4x - 2x - 24 = 0
x^2 + 2x = 24 = 0
Factors to
(x+6)(x-4) = 0
The positive solution
x = 4 is the youngest, I'll let you find the ages of the other two
Check your answers in the given statement
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