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Math Help - Quadratic Equation Word Problems

  1. #1
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    Quadratic Equation Word Problems

    Hello

    Does anyone know how to figure out these equations using the quadratic formula? Thanks a lot!!!

    1) Find two consecutive positive integers whose product is 272
    (I know how to find the integers I just don't understand how to put this into a quadratic equation and find the answer?)

    2)A rectangular piece of sheet metal has a length that is twice the width. In each corner, a square of side 3in is cut out, and the outer strips are then bent up to form an open box with volume 168in cubed. Find the dimesions of the original sheet

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  2. #2
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    Quote Originally Posted by stephie View Post
    Hello

    Does anyone know how to figure out these equations using the quadratic formula? Thanks a lot!!!

    1) Find two consecutive positive integers whose product is 272
    (I know how to find the integers I just don't understand how to put this into a quadratic equation and find the answer?)
    Let the first integer be x. Then the next consecutive integer is x + 1. Thus:
    x(x + 1) = 272

    x^2 + x - 272 = 0

    Thus
    x = (-1 (+/-) sqrt(1^2 - 4*1*(-272)))/(2*1)

    x = (-1 (+/-) sqrt(1 + 1088))/2

    x = (-1 (+/-) 33)/2

    Thus x = 16 or x = -17

    So our two consecutive integers are 16 and 17 or -17 and -16.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by stephie View Post
    Hello

    Does anyone know how to figure out these equations using the quadratic formula? Thanks a lot!!!

    2)A rectangular piece of sheet metal has a length that is twice the width. In each corner, a square of side 3in is cut out, and the outer strips are then bent up to form an open box with volume 168in cubed. Find the dimesions of the original sheet

    Let the length of the sheet be L and the width W.

    We are cutting out a 3 x 3 square in each corner, so the dimensions of the box will be (L - 3) x (W - 3) x 3. The volume of this box is thus:
    V = 3(L - 3)(W - 3) = 168

    Since L = 2W this becomes:
    V = 3(2W - 3)(W - 3) = 3(2W^2 - 9W + 9) = 6W^2 - 27W + 27 = 168

    So
    6W^2 - 27W -141 = 0

    W = (-(-27) (+/-) sqrt((-27)^2 - 4*6*(-414)))/(2*6)

    W = (27 (+/-) sqrt(729 + 3384))/12

    W = (27 (+/-) sqrt(4113))/12

    W = (27 (+/-) 3sqrt(457))/12

    W = (9 (+/-) sqrt(457))/4

    Now sqrt(457) is about 21.3776 and this is greater than 9, so the "-" solution comes out to be negative. As we can't have a negative width, we discard this possibility. Thus the only solution is the "+" solution:
    W = (9 + sqrt(457))/4 (approx. 7.59439)

    And
    L = 2W = (9 + sqrt(457))/2 (approx. 15.1888)

    -Dan
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    You're awesome! Thanks so much
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    Forum Admin topsquark's Avatar
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    You're very welcome! Glad to help.

    -Dan
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