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Math Help - Need help quadradic equations

  1. #1
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    Need help quadradic equations

    Solve using the quadratic formula

    x^2-7x+12=0

     2x^2 - 7x - 4 = 0


    Use the zero product properity to solve for x and k

    6k^2-k-1=0
    x^2+6x-7=0

    I hate posting this many problems but the math class i'm in, I can't learn, its an overcrowded classroom with people who are there that don't care to learn and just do whatever, I'm getting my mom to call so I can get switched out of the class and put into one that i'll have the ability to learn more, anyways, thank you for any help that anyone is able to give.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Haxcake View Post
    Solve using the quadratic formula

    x^2-7x+12=0

     2x^2 - 7x - 4 = 0
    The quadratic formula says:
    Given ax^2 + bx + c = 0 then x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    So for the first one:
    x^2-7x+12=0

    a = 1, b = -7, c = 12

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

    x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(12)}}{2 \cdot 1}

    x = \frac{7 \pm \sqrt{49 - 48}}{2}

    x = \frac{7 \pm \sqrt{1}}{2}

    x = \frac{7 \pm 1}{2}

    So x = \frac{7+1}{2} = 4 or x = \frac{7 - 1}{2} = 3.

    I'll let you do the second one. I got x = 4 or x = -1/2.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Haxcake View Post
    Use the zero product properity to solve for x and k

    6k^2-k-1=0
    x^2+6x-7=0
    This uses the property of real numbers:
    If ab = 0 then either a = 0 or b = 0 (or both).

    The rest of it is factoring.

    The first one says:
    6k^2-k-1=0

    (2k - 1)(3k + 1) = 0 (If you need help with factoring, just let me know.

    So either 2k - 1 = 0 ==> k = 1/2
    or
    3k + 1 = 0 ==> k = -1/3.

    Again, I'll leave the other one to you. I get x = 1 or x = -7.

    -Dan
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  4. #4
    Eater of Worlds
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    The quadratic formula can be looked up anywhere.

    As for the second half of your post:

    I reckon by 'zero product property', they mean factoring?. If so, why don't they just say so.
    If I am correct, why would they call it that except to show off and act pretentious.

    Factor:

    6k^{2}-k-1

    Because the leading coefficient is 6, we must compensate and find

    what numbers add to -1 and when multiplied equal -6.

    That would be 2 and -3, wouldn't it?.

    6k^{2}+2k-3k-1

    (6k^{2}+2k)-(3k+1)

    Factor:

    2k(3k+1)-(3k+1)

    (2k-1)(3k+1)

    I hope this is what they mean by zero product property.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    x^{2}+6x-7

    Do the same thing:

    What 2 numbers when added equal 6 and when multiplied equal -7

    That'd be 7 and -1.

    x^{2}-x+7x-7

    (x^{2}-x)+(7x-7)

    Factor:

    x(x-1)+7(x-1)

    (x-1)(x+7)
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  5. #5
    Sophie_15
    Guest

    Smile i am doin the same thing i could only get the first one and the last on tough.

    Quote Originally Posted by Haxcake View Post
    Solve using the quadratic formula

    x^2-7x+12=0

     2x^2 - 7x - 4 = 0


    Use the zero product properity to solve for x and k

    6k^2-k-1=0
    x^2+6x-7=0

    I hate posting this many problems but the math class i'm in, I can't learn, its an overcrowded classroom with people who are there that don't care to learn and just do whatever, I'm getting my mom to call so I can get switched out of the class and put into one that i'll have the ability to learn more, anyways, thank you for any help that anyone is able to give.
    1. x^2-7x+12=0
    (x-3)(x-4)
    x-3=0 x=4
    x-4=0 x=3

    4. x^2+6x-7=0
    (x+7)(x-1)
    x+7=0 x=1
    x-1=0 x=-7
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