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Math Help - [SOLVED] equation help!!!!

  1. #1
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    [SOLVED] equation help!!!!

    im in a review section in my class. this is some stuff i havent done in yrs. so if you could help with these 2 problems thatd be great (if possible, can you explain how you got the answer? i really dont remember this from that far back). . .

    1. Find the value of K such that the quadratic equation x^2+(2K+1)x-(K/4) = 0 has only one solution


    2.Determine the value(s) of x that makes the following statement true: 6x^(4/3) + 7x(2/3) = 3


    plz & thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by liltiger22 View Post
    im in a review section in my class. this is some stuff i havent done in yrs. so if you could help with these 2 problems thatd be great (if possible, can you explain how you got the answer? i really dont remember this from that far back). . .

    1. Find the value of K such that the quadratic equation x^2+(2K+1)x-(K/4) = 0 has only one solution
    A quadratic equation only has one solution when the discriminant is equal to zero.

    Note that the discriminant is defined by \Delta=\sqrt{b^2-4ac}

    Since our quadratic equation has the form ax^2+bx+c=0, we see that the discriminant then would be \sqrt{(2k+1)^2+4(1)(K/4)}

    Set this equal to zero and solve for K.

    Can you try to take it from here?

    2.Determine the value(s) of x that makes the following statement true: 6x^(4/3) + 7x(2/3) = 3


    plz & thank you
    With a little rearrangement, we see that this equation can be written as 6(x^{\frac{2}{3}})^2+7x^{\frac{2}{3}}=3

    Now make a little substitution: let's say z=x^{\frac{2}{3}}

    Then this becomes a quadratic equation: 6z^2+7z=3\implies 6z^2+7z-3=0

    Solve this equation, and then back substitute z=x^{\frac{2}{3}} to solve for x.

    Can you try to take it from here?

    Does this make sense?

    --Chris
    Last edited by Chris L T521; October 5th 2008 at 07:53 PM. Reason: typo
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  3. #3
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    1. In order for the quadratic equation ax^2 + bx + c = 0 to have only one solution, the discriminant b^2 - 4ac must be equal to zero.

    The discriminant for x^2 + (2k+1)x - k/4 = 0 is (2k+1)^2 - 4(1)(-k/4) = (2k + 1)^2 + k.

    Hence (2k + 1)^2 + k = 0

    4k^2 + 4k + 1 + k = 0

    4k^2 + 5k + 1 = 0

    (4k + 1)(k + 1) = 0

    so k = -1 or k = -1/4.
    Last edited by icemanfan; October 5th 2008 at 07:55 PM.
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  4. #4
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    Quote Originally Posted by icemanfan View Post
    1. In order for the quadratic equation ax^2 + bx + c = 0 to have only one solution, the discriminant b^2 - 4ac must be equal to zero.

    The discriminant for x^2 + (2k+1)x - k/4 = 0 is (2k+1)^2 - 4(1)(k/4) = (2k + 1)^2 - k.

    Hence (2k + 1)^2 - k = 0

    4k^2 + 4k + 1 - k = 0

    4k^2 + 3k + 1 = 0

    k = \frac{-3 \pm \sqrt{3^2 - 4(4)(1)}}{8}

    so there are no real solutions.
    I don't want to pick at you but shouldn't it be:

    x^2 + (2k+1)x - k/4 = 0 is (2k+1)^2 - 4(1)(\bold{{\color{red}-}}k/4) = (2k + 1)^2\bold{{\color{red}+}} k.
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  5. #5
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    Quote Originally Posted by earboth View Post
    I don't want to pick at you but shouldn't it be:

    x^2 + (2k+1)x - k/4 = 0 is (2k+1)^2 - 4(1)(\bold{{\color{red}-}}k/4) = (2k + 1)^2\bold{{\color{red}+}} k.
    You're right. I'm going to edit my post now.
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  6. #6
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    Exclamation

    well i got number 2, but im still workin on number 1 . . . but if you could help on number 1, thatd be great
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