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Thread: Someone please help solve this problem

  1. #1
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    Someone please help solve this problem

    A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.
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  2. #2
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    Quote Originally Posted by moomoo762 View Post
    A rectangular box has sides with lengths x, x+1, and x+2. If the volume of the box is 504 cubic inches, find the dimensions of the box.
    $\displaystyle x(x+1)(x+2) = 504$

    note the prime factorization of 504 ...

    $\displaystyle 504 = 7 \cdot 2^3 \cdot 3^2$
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  3. #3
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    You should know that the formula for the volume of a rectangular prism is
    V = LWH.

    So plug in:
    $\displaystyle 504 = x(x + 1)(x + 2)$

    (let's switch sides)
    $\displaystyle x(x + 1)(x + 2) = 504$

    Multiply out the left side, and afterwards, subtract 504 from both sides. You'll end up with a cubic equation equaling zero. Solve for x. Don't forget to reject any solution that gives you a negative number for any of the lengths.


    01


    EDIT: beaten to it by skeeter!
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  4. #4
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    I'm stuck

    I subtract 504 from both sides to make it equal to zero and I get
    x^3+5x^2+2x-504=0

    What's the best way to solve for x now?
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  5. #5
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    Quote Originally Posted by moomoo762 View Post
    I'm stuck

    I subtract 504 from both sides to make it equal to zero and I get
    x^3+5x^2+2x-504=0

    What's the best way to solve for x now?
    rational root theorem
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  6. #6
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    Using the rational root test I got 7

    I know the other two sides are 8 and 9 but how do I go about finding them?
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  7. #7
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    Quote Originally Posted by moomoo762 View Post
    Using the rational root test I got 7

    I know the other two sides are 8 and 9 but how do I go about finding them?
    first of all, correct your expansion ...

    $\displaystyle x(x+1)(x+2) = 504$

    $\displaystyle x^3 + 3x^2 + 2x - 504 = 0$

    you know 7 is a root ... use synthetic division

    Code:
    7]  1.......3........2.......-504
    ............7.......70........504
    -------------------------------
    ....1......10.......72.........0
    note the coefficients of the depressed polynomial ...

    $\displaystyle x^2 + 10x + 72 = 0$

    since $\displaystyle b^2-4ac < 0$ , the other two roots are imaginary.

    $\displaystyle x = 7$ is the only real root

    go back to the original equation ...

    $\displaystyle x(x+1)(x+2) = 504$

    $\displaystyle 7(7+1)(7+2) = 504$
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  8. #8
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    $\displaystyle x(x+1)(x+2) = 504$

    $\displaystyle x(x^2+3x+2)=504$

    $\displaystyle x^3+3x^2+2x-504=0$

    $\displaystyle (x-7)(x^2+10x+72)=0$

    Notice that there are no real solutions for the quadratic factor as $\displaystyle \triangle <0$

    So the only real solution is $\displaystyle x=7$

    The question stated the sides are $\displaystyle x,\ x+1,\ x+2$ so one side is $\displaystyle 7$ and the others are $\displaystyle 7+1=8$ and $\displaystyle 7+2=9$
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