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Math Help - solving equation

  1. #1
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    solving equation

    Greetings all,
    I have a test tomorrow and i am having difficulties with one section of the study guide. If anyone could help me with solutions that would be really appreciated.

    Basically he wants us to find all solutions and factor the following problems.

    1. 54x + 2x^3=0

    2. -11 + 2x - 3x^2=0

    3. 3x^4 + 18x^2 - 48=0

    4. x^3-8=0


    Once again thank you!!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    1.  x(54+2x^2) = 0 so  x = 0 or  x^2 = -26 \implies x = \pm i \sqrt{26} .

    2.  -3x^2 +2x - 11 . Use quadratic formula.

    3.
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  3. #3
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    3. 3x^{4} + 18x^{2} - 48=0
    Divide by 3 and get x^{4}+6x^{2}-16=0

    Let u=x^{2}

    u^{2}+6u-16=0

    (u-2)(u+8)=0

    3(x^{2}-2)(x^{2}+8)=0

    4. x^{3}-8=0
    This is just the difference of two cubes


    Once again thank you!! [/quote]
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by WolfMV View Post
    4. x^3-8=0
    Let me expand a little on galactus' comment, since most who haven't seen how to solve this would likely miss two solutions.

    You need to solve
    x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Let me expand a little on galactus' comment, since most who haven't seen how to solve this would likely miss two solutions.

    You need to solve
    x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0

    -Dan
    How's this complicated?

    x^3 = 8, and then it's easy to see x = 2.

    I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

    Did I miss something?
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  6. #6
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    Quote Originally Posted by Ideasman View Post
    How's this complicated?

    x^3 = 8, and then it's easy to see x = 2.

    I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

    Did I miss something?
    The second factor is never equal to zero when its is real. It can still have complex solutions.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    How's this complicated?

    x^3 = 8, and then it's easy to see x = 2.

    I see that x^3 - 8 factors into (x - 2)(x^2 + 2x + 4); this is equal to 0 when x = 2 for the (x - 2) but for the (x^2 + 2x + 4) factor, this will never equal 0.

    Did I miss something?
    Use the quadratic formula on the quadratic factor.

    -Dan
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