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Math Help - finding real solutions of these examples

  1. #1
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    finding real solutions of these examples

    lets try this latex thing out...

    3x^4-2x^2-1=0

    I know I need to move the one to the other side, then I think factor our what is common, in this case.. it would be a x^2 right? so I get to

    x^2(3x^2-2)=1

    I get stumped after that.

    Another one I have is
    x^6+7x^3-8=0
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Geomatt View Post
    lets try this latex thing out...

    3x^4-2x^2-1=0

    I know I need to move the one to the other side, then I think factor our what is common, in this case.. it would be a x^2 right? so I get to

    x^2(3x^2-2)=1

    I get stumped after that.

    Another one I have is
    x^6+7x^3-8=0
    The first is quadratic in x^2, the second is quadratic in x^3.


    Note that:

    3x^4 - 2x^2 - 1 = 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0

    similarly,

    x^6 + 7x^3 - 8 = \left( x^3 \right)^2 + 7 \left( x^3 \right) - 8 = 0

    can you take it from here?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Note that:

    3x^4 - 2x^2 - 1 = 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0

    similarly,

    x^6 + 7x^3 - 8 = \left( x^3 \right)^2 + 7 \left( x^3 \right) - 8

    can you take it from here?
    lets see, so it would be 3x-2x^2-1=0 which I could then rearrange and solve with the quad. formula?

    im not sure about the second example.


    Im sorry if the answer is obvious, math has always been tough to me.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Geomatt View Post
    lets see, so it would be 3x-2x^2-1=0 which I could then rearrange and solve with the quad. formula?

    im not sure about the second example.


    Im sorry if the answer is obvious, math has always been tough to me.
    for the first:

    how would you solve 3y^2 - 2y - 1 = 0? what is the similarity between this last equation and the one i gave you for the first?
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    for the first:

    how would you solve 3y^2 - 2y - 1 = 0? what is the similarity between this last equation and the one i gave you for the first?
    ok so
    the first example
    3x-2x^2-1=0
    2x^2-3x+1=0
    (2x-1)(x-1)
    x= -1/2, x=1

    I checked my answers and im supposed to get x=1, -1, i messed up somewhere.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Geomatt View Post
    ok so
    the first example
    3x-2x^2-1=0
    2x^2-3x+1=0
    (2x-1)(x-1)
    x= -1/2, x=1

    I checked my answers and im supposed to get x=1, -1, i messed up somewhere.
    this is wrong because you have the wrong question. it is 3x^4 - 2x^2 - 1 = 0 you forgot the fourth power

    i'll do the first, you do the second

    3x^4 - 2x^2 - 1 = 0 ......rewrite

    \Rightarrow 3 \left( x^2 \right)^2 - 2 \left( x^2 \right) - 1 = 0 ........now factor as if the x^2 was just a variable by itself, like y or something

    \Rightarrow \left(3x^2 + 1 \right) \left( x^2 - 1 \right) = 0

    \Rightarrow x^2 = - \frac {1}{3} \mbox { or } x^2 = 1

    \Rightarrow x = \sqrt { \frac {1}{3}}i \mbox { or } x = \pm 1


    you don't want the complex roots, so just take the  \pm 1 answer, that's the real roots solution
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    \Rightarrow x^2 = - \frac {1}{3} \mbox { or } x^2 = 1

    \Rightarrow x = \sqrt { \frac {1}{3}}i \mbox { or } x = \pm 1


    \Rightarrow x = \pm \sqrt { \frac {1}{3}}~i \mbox { or } x = \pm 1

    RonL
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    \Rightarrow x = \pm \sqrt { \frac {1}{3}}~i \mbox { or } x = \pm 1

    RonL
    yeah, that's what i meant
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