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Math Help - Quadratic equation help

  1. #1
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    Quadratic equation help

    Hi,

    I have come up against a question in a past paper I am revising for a maths exam next week and I am having some difficulty.

    The question is:

    A concrete path with uniform width x metres is to be built around the rectangular base of a house of length 61 metres and width 35 metres. The area of the path is 400m^2. Set up a quadratic equation in x and solve it to give the width of the path.

    So far I have:

    (61+2x)(35+2x)-(61*35)=400

    (2135+122x+70x+4x^2)-(2135)=400

    192x+4x^2=400

    192x+4x^2-400=0

    48x+x^2-100=0
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  2. #2
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    Quote Originally Posted by cjmcdougall View Post
    Hi,

    I have come up against a question in a past paper I am revising for a maths exam next week and I am having some difficulty.

    The question is:

    A concrete path with uniform width x metres is to be built around the rectangular base of a house of length 61 metres and width 35 metres. The area of the path is 400m^2. Set up a quadratic equation in x and solve it to give the width of the path.

    So far I have:

    (61+2x)(35+2x)-(61*35)=400

    (2135+122x+70x+4x^2)-(2135)=400

    192x+4x^2=400

    192x+4x^2-400=0

    48x+x^2-100=0
    => (x + 50)(x - 2) = 0 ....
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  3. #3
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    Hello, cjmcdougall

    A concrete path with uniform width \,x metres is to be built around
    the rectangular base of a house of length 61 m and width 35 m.
    The area of the path is 400 m^2.
    Set up a quadratic equation in \,x and solve it to give the width of the path.

    So far I have:

    . . (61 + 2x)(35 + 2x) - (61)(35) .= .400

    . . 2135 + 122x + 70x + 4x^2 - 2135 .= .400

    . . 2135 + 122x + 70x + 4x^2 - 2135 .= .400

    . . 192x + 4x^2 .= .400

    . . 192x + 4x^2 - 400 .= .0

    . . 48x + x^2 - 100 .= .0


    And you don't know how to solve a quadratic equation?
    . . Then why was this problem assigned to you?


    You have: .x^2 + 48x - 100 = 0

    Factor: .(x - 2)(x + 50) = 0

    And we have: .x = 2, x = -50


    The only realistic answer is: .x = 2 meters.

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  4. #4
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    I was assigned this as on my course I am expected to already know this even though I have never been taught it.
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  5. #5
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    Quote Originally Posted by cjmcdougall View Post
    I was assigned this as on my course I am expected to already know this even though I have never been taught it.
    In which case, you could consider another way
    to lead you into solving quadratics....

    Draw a diagram of the path surrounding the house
    Continue all horizontal and vertical lines from the inner rectangle to the outer rectangle.

    x^2+61x+x^2+35x+x^2+61x+x^2+35x = 400
    2x^2+61x+35x = 200

    2x^2+96x = 200

    x^2+48x = 100

    x(x+48) = 100

    one of the factors is 48 more than the other.
    Find the factors of 100 that differ by 48.

    100 = 100(1)
    100 = 50(2)

    etc.
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  6. #6
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    Quote Originally Posted by cjmcdougall View Post
    I was assigned this as on my course I am expected to already know this even though I have never been taught it.
    What would your teacher say to that?
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  7. #7
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    Quote Originally Posted by Wilmer View Post
    What would your teacher say to that?
    I'm coaching two-thirds of the students from a class,
    whose teacher has told them he will not be covering
    a quarter of their entire syllabus.
    The course is 2 years in duration.
    It's quite common for teachers not to cover the syllabus in time.
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  8. #8
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    Then you should talk to the chair of the department about that. The printed syllabus for a class, or description in the school catalog, is a legal contract. While a certain amount of "contraction" is acceptable, any school administrator, including chair of the department, should be very sensitive to the possiblility of a law suit. If a teacher covers only 3/4 of the syllabus, repeatedly, that is a very dangerous situation.
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