Question 115286: find two consecutive odd integers whose product is 15 more than three times their sum.
My guess is x(x+2)=3x+15, when I work this problem I have to use the quadratic formula to get an answer I don't think is correct, 1/2 +- square root of 15.
Thank you
LeeAnn
Found 3 solutions by stanbon, ganesh, MathLover1:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! find two consecutive odd integers whose product is 15 more than three times their sum.
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Odd integers are always one more (or less) than even integers.
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1st odd integer: 2x+1
next odd integer:2x+3
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EQUATION:
(2x+1)(2x+3)= 3[(2x+1)+(2x+3)] + 15
4x^2 + 8x + 3 = 12x+27
4x^2 -4x -24 = 0
x^2-x-6 = 0
(x-3)(x+2) = 0
x = 3 or x=-2
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If x=3 you get:
2x+1 = 7 and 2x+3 = 9
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If x = -2 you get:
2x+1 = -3 and 2x+3 = -1
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Cheers,
Stan H.
Answer by ganesh(20) (Show Source):
You can put this solution on YOUR website! Let x and x + 2 be two consecutive odd integers.
Their product is x (x+2)= x^2 + 2x.
Their sum is x + x + 2 = 2x + 2.
Three times their sum = 3 (2x + 2) = 6x + 6.
Given that, their product is 15 more than three times their sum.
This can be mathematically written as x^2 + 2x = 6x + 6 + 15.
Simplifying, we get x^2 - 4x -21 = 0.
Solve.
x^2 - 7x + 3x - 21 = 0
Or, x(x - 7) + 3(x - 7) = 0
Or, (x - 7) (x + 3) = 0
The solutions are x = 7, -3.
So, -3, -1 is a set of solution and 7,9 is another set of solution.
Answer by MathLover1(20850)  (Show Source):
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