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Math Help - Need help on these 3 problems

  1. #1
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    Angry Need help on these 3 problems

    I really need help with these.



    Solve the equation for x. (If there is one solution with a multiplicity of two, enter it in both answer boxes.)
    a^2x^2 + 8ax + 16 = 0
    i think there is a larger and smaller value



    Find all values of k that ensure that the given equation has exactly one solution.
    6x^2 + kx + 9 = 0

    there is a smaller and larger value


    A wire 370 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures have the same area, what are the lengths of the two pieces of wire?
    No idea on this one =(
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Fuze View Post
    I really need help with these.



    Solve the equation for x. (If there is one solution with a multiplicity of two, enter it in both answer boxes.)
    a^2x^2 + 8ax + 16 = 0
    i think there is a larger and smaller value
    use the quadratic formula


    Find all values of k that ensure that the given equation has exactly one solution.
    6x^2 + kx + 9 = 0
    a quadratic has one root if it's discriminant is zero.

    the discriminant for a quadratic of the form y = ax^2 + bx + c is given by \triangle = b^2 - 4ac

    A wire 370 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures have the same area, what are the lengths of the two pieces of wire?
    No idea on this one =(
    Let x be the length that is cut to make the square. then the length to make the circle will be 370 - x (this will be the circumference of the circle).

    the side-length of the square will be \frac x4 and we can use the formula for the circumference of a circle to find the radius of a circle whose circumference is 370 - x. thus we can find formulas for the areas of the two figures. now we just equate them and solve for x

    (if you wish, you can let the length for the square be 4x, this avoids the awkwardness of x/4, but you should note that when you solve for x, you must multiply the value by 4 again to get the answer for the length of this piece)
    Last edited by Jhevon; January 13th 2008 at 10:04 PM. Reason: corrected an error (thanks TwistedOne)
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Fuze View Post
    I really need help with these.

    Solve the equation for x. (If there is one solution with a multiplicity of two, enter it in both answer boxes.)
    a^2x^2 + 8ax + 16 = 0
    i think there is a larger and smaller value
    Use the quadratic formula.

    Quote Originally Posted by Fuze View Post

    Find all values of k that ensure that the given equation has exactly one solution.
    6x^2 + kx + 9 = 0

    there is a smaller and larger value
    Delta = 0

    b^2 - 4ac = 0

    k^2 - 4(6)(9) = 0

    k^2 = 216

    k = 6 \sqrt{6}

    Quote Originally Posted by Fuze View Post
    A wire 370 in. long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures have the same area, what are the lengths of the two pieces of wire?
    No idea on this one =(
    One wire length = x
    The other = 370 - x

    Perimeter of square = 4L

    Area of square = L^2

    So 4L = 370 - x

    Then L = \frac{370 - x}{4}

     L^2 = \frac{(370 - x)^2}{16}

    Perimeter of circle = 2 \pi r
    Area of a circle = \pi r^2

    If 2 \pi r = x then:

    r = \frac{x}{2 \pi}

    Area of the circle  = \pi \left( \frac{x^2}{4 (\pi)^2} \right)

    If the area is the same then  L^2 = \pi r^2

    And you can take it from here.
    Last edited by janvdl; January 13th 2008 at 09:56 PM. Reason: darn latex :-D
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  4. #4
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    Correction

    On the second problem:
    There are actually two different values for k that solve the problem:
    k=\pm6\sqrt{6}

    --Kevin C.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    On the second problem:
    There are actually two different values for k that solve the problem:
    k=\pm6\sqrt{6}

    --Kevin C.
    Indeed! don't i look stupid!

    thanks for the catch Kevin
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  6. #6
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    thanks guys! really helped me alot
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