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Math Help - Need Quadratic Equation

  1. #1
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    Need Quadratic Equation

    Just need a little boost to solve this question..

    " The Parks department is planning a new flower bed outside city hall. It will be rectangular with dimensions by 9 m by 6m. The flower bed will be surrounded by a path of constant width with the same area as the flower bed. Calculate the perimeter of the path."

    Diagram attatched.


    A) Set up a quadratic Equation To model the problem
    BTW, Grade 10 Math.
    Attached Thumbnails Attached Thumbnails Need Quadratic Equation-sdfdsf.jpg  
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  2. #2
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    Quote Originally Posted by letsgofam View Post
    Just need a little boost to solve this question..

    " The Parks department is planning a new flower bed outside city hall. It will be rectangular with dimensions by 9 m by 6m. The flower bed will be surrounded by a path of constant width with the same area as the flower bed. Calculate the perimeter of the path."

    Diagram attatched.


    A) Set up a quadratic Equation To model the problem
    BTW, Grade 10 Math.
    (2x+6)(2x+9) = 2(54)

    solve for x, then determine the path's perimeter
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  3. #3
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    Quote Originally Posted by skeeter View Post
    (2x+6)(2x+9) = 2(54)

    solve for x, then determine the path's perimeter
    Alright
    So
    (2x+6)(2x+9) = 2(54)
    I simplify that
    Get

    4x^2 + 30x +54 = 108
    4x^2 + 30x -54=0

    Apply the quadratic formula, solve for x ?
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  4. #4
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    Okay, I got -3 and -4.5 for the the values of X, but dimensions can't be negative numbers , what's wrong?
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by letsgofam View Post
    Okay, I got -3 and -4.5 for the the values of X, but dimensions can't be negative numbers , what's wrong?
    your answers for the values of x are wrong.. calculation error.. re do your quadratic equation again
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  6. #6
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    Quote Originally Posted by harish21 View Post
    your answers for the values of x are wrong.. calculation error.. re do your quadratic equation again
    How?

    2x + 6 = 0
    2x = =-6
    x= -3

    2x+9 = 0
    2x = -9
    x = -4.5

    ??
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by letsgofam View Post
    How?

    2x + 6 = 0
    2x = =-6
    x= -3

    2x+9 = 0
    2x = -9
    x = -4.5

    ??
    You are assuming that (2x + 6)(2x+9)=0 This is NOT true in your case..

    Look at what you have done in post number 3 of this thread.. you obtained a quadratic equation:4x^2 + 30x -54=0

    this is where you need to find your x from.

    An unknown x in a qudratic equation ax^2+bx+c (where a,b,c,are known) can be calculated as:

    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}


    now compare the equation you obtained 4x^2 + 30x -54=0 with the above formula and find x.
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  8. #8
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    Great, I solved the equation.

    Got 9M for the widths, and 12 for the lengths.
    Thanks alot guys!
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