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Math Help - what's x's value?

  1. #1
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    what's x's value?

    could you tell me what is X's Value?
    Attached Thumbnails Attached Thumbnails what's x's value?-x-4-x-3-4.gif  
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neworld222 View Post
    could you tell me what is X's Value?
    Yikes!
    x^4 - x^3 - 4 = 0
    can be done analytically, but is much easier to approximate numerically. I get
    x = -1.21733 or x = 1.7484
    as the only real solutions.

    -Dan
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Yikes!
    x^4 - x^3 - 4 = 0
    can be done analytically, but is much easier to approximate numerically. I get
    x = -1.21733 or x = 1.7484
    as the only real solutions.

    -Dan
    those are correct, i used a utility on my pc to find them

    how did you approximate it numerically (i was thinking about using Newton-Raphson approximation, but i couldn't bother)?

    how can it be done analytically?
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  4. #4
    Eater of Worlds
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    Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

    The "Quartic Formula"
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    Quote Originally Posted by galactus View Post
    Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

    The "Quartic Formula"
    thx,though the method is funny,but i think it's the best one if i don't want solve this subject by use advanced-algebra
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by neworld222 View Post
    thx,though the method is funny,but i think it's the best one if i don't want solve this subject by use a advanced-algebra
    well, that's true i guess. Newton-Raphson approximation is a Calc 1 topic, so it's not really that advanced... in the way we would think of advanced.

    the quartic formula, which is so cumbersome that i don't see my self using it anytime soon (if ever), is probably the method your advanced algebra professor would want. don't blame me if it takes up 6 pages (back and front) though
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    well, that's true i guess. Newton-Raphson approximation is a Calc 1 topic, so it's not really that advanced... in the way we would think of advanced.

    the quartic formula, which is so cumbersome that i don't see my self using it anytime soon (if ever), is probably the method your advanced algebra professor would want. don't blame me if it takes up 6 pages (back and front) though
    i should'nt blame you,if i'll have a headache when i calculate by 'advanced method',to a certainty blame Mr.Newton
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by galactus View Post
    Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

    The "Quartic Formula"
    Thank you! This is the answer to a post I made a while ago. (I was looking for the logic behind the quartic formula.)

    -Dan
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  9. #9
    Eater of Worlds
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    Thanks for the thanks. How serendipitous
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  10. #10
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    Certain Computer Algebra Systems can solve certain equations, for example, Mathematica.

    This is what I got (not exactly sure what it means because it says "too long, it will appear as a clipboard").

    Code:
    MATRIX([[1/4+1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)
    +576)/(54+6*3153^(1/2))^(1/3))^(1/2)+1/12*6^(1/2)*((3*(54+6*3153^(1/2))^(1/3)*
    ((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^
    (1/3))^(1/2)+2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+
    6*3153^(1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)-96*((9*(54+6*3153^(1/2))^(
    1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*
    3153^(1/2))^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6
    *3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(1/2)], [1/4+1/12*((9*(54
     +6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))
    ^(1/2)-1/12*6^(1/2)*((3*(54+6*3153^(1/2))^(1/3)*((9*(54+6*3153^(1/2))^(1/3)-12*(
    54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+2*((9*(54+6*3153^(
    1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)*(54+
    6*3153^(1/2))^(2/3)-96*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+
    576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*3153^(1/2))^(1/3))/(54+6*3153^(
    1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*
    3153^(1/2))^(1/3))^(1/2))^(1/2)], [1/4-1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*
    3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+1/12*I*6^(1/2)*((-3*(54+6
    *3153^(1/2))^(1/3)*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/
    (54+6*3153^(1/2))^(1/3))^(1/2)-2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(
    1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)+96*((9*
    (54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(
    1/3))^(1/2)+9*(54+6*3153^(1/2))^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(
    1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(
    1/2)], [1/4-1/12*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54
    +6*3153^(1/2))^(1/3))^(1/2)-1/12*I*6^(1/2)*((-3*(54+6*3153^(1/2))^(1/3)*((9*(54+
    6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(
    1/2)-2*((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^
    (1/2))^(1/3))^(1/2)*(54+6*3153^(1/2))^(2/3)+96*((9*(54+6*3153^(1/2))^(1/3)-12*(
    54+6*3153^(1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2)+9*(54+6*3153^(1/2))
    ^(1/3))/(54+6*3153^(1/2))^(1/3)/((9*(54+6*3153^(1/2))^(1/3)-12*(54+6*3153^(
    1/2))^(2/3)+576)/(54+6*3153^(1/2))^(1/3))^(1/2))^(1/2)]])
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  11. #11
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    Quote Originally Posted by galactus View Post
    Solving a quartic can be done, but it is cumbersome. Here is a nice step-by-step approach to the process.

    The "Quartic Formula"
    Personally, I prefer The "TI-89 Formula"

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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    Certain Computer Algebra Systems can solve certain equations, for example, Mathematica.

    This is what I got (not exactly sure what it means because it says "too long, it will appear as a clipboard").
    In this case Ma is making a meal of it. Looking only for the real roots derive gives what is in the attachment:

    RonL
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