Question 551087: Find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7. Times the second
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7. Times the second
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"Find three consecutive positive odd integers"
x, (x+2), (x+4)
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such that the product of the first and the third is 4 less than 7. Times the second
x(x+4) = 7(x+2) - 4
x^2 + 4x = 7x + 14 - 4
x^2 + 4x = 7x + 10
Combine like terms on the left
x^2 + 4x - 7x - 10 = 0
x^2 - 3x - 10 = 0
Factors to
(x-5)(x+2) = 0
two solutions
x = 5, the positive odd integer, so x=5 is the only solution
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the three integers, 5, 7, 9
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see if that works in the statement:
"Find three consecutive positive odd integers such that the product of the first and the third is 4 less than 7. Times the second"
5*9 = 7(7) - 4
45 = 49 - 4; confirms our solution
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