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Math Help - Solving quadratic equation with surds

  1. #1
    Newbie Miasmagasma's Avatar
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    Solving quadratic equation with surds

    solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

    i can't isolate the '6√3' without getting something wrong and ruining the equation.
    Last edited by mr fantastic; December 9th 2009 at 06:29 PM. Reason: Moved from existing thread
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  2. #2
    Senior Member Stroodle's Avatar
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    Hi there,

    Try rearranging it into the form:

    (1-6\sqrt{3})x^2+24=0

    Can you take it from there?
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Miasmagasma View Post
    solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

    i can't isolate the '6√3' without getting something wrong and ruining the equation.
    Are you sure this is the right equation? Is the second coeffiecent to the first degree?
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  4. #4
    MHF Contributor
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    I would suggest using the Quadratic Formula...
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  5. #5
    Newbie Miasmagasma's Avatar
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    wow i'm even more confused now

    you have to be gentle with me as i'm teaching myself this stuff.

    i can't see how to get the equation to that point, stroodle. and i wouldn't know what to do from there anyway.

    ??
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  6. #6
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Are you sure this is the right equation? Is the second coeffiecent to the first degree?
    i'm afraid i don't understand what you mean.
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  7. #7
    Senior Member Stroodle's Avatar
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    If you didn't mean to square the x in the second term of the equation;

    i.e. x^2-(6\sqrt{3})x+24=0

    Then you could either use the quadratic formula or complete the square (as follows) to solve for x

    x^2-(6\sqrt{3})x+24=0

    \left (x-\frac{6\sqrt{3}}{2}\right ) ^2-\frac{108}{4}+\frac{96}{4}=0

    \left ( x-\frac{6\sqrt{3}}{2}+\frac{2\sqrt{3}}{2}\right )\left ( x-\frac{6\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right )=0

    (x-2\sqrt{3})(x-4\sqrt{3})=0

    So, x=2\sqrt{3} or  x=4\sqrt{3}

    Hope that helps
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  8. #8
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by Miasmagasma View Post
    wow i'm even more confused now

    you have to be gentle with me as i'm teaching myself this stuff.

    i can't see how to get the equation to that point, stroodle. and i wouldn't know what to do from there anyway.

    ??
    If the equation is x^2-(6\sqrt{3})x^2+24=0

    Then you can rearrange it to the form:

    (1-6\sqrt{3})x^2+24=0 ,as when you expand (1-6\sqrt{3})x^2 you get x^2-(6\sqrt{3})x^2

    From there you can simply solve as follows:

    (1-6\sqrt{3})x^2+24=0

    (1-6\sqrt{3})x^2=-24

    x^2=\frac{-24}{1-6\sqrt{3}}

    x=\pm\sqrt{\frac{24}{6\sqrt{3}-1}}
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  9. #9
    MHF Contributor
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    Even if it's not, you can still use the quadratic formula anyway.
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  10. #10
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Prove It View Post
    Even if it's not, you can still use the quadratic formula anyway.

    as Prove it said you could use the quad equation letting

    A= (1-6\sqrt{3})

    B= 0

    C = 24
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  11. #11
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by Stroodle View Post
    If you didn't mean to square the x in the second term of the equation;

    i.e. x^2-(6\sqrt{3})x+24=0

    Then you could either use the quadratic formula or complete the square (as follows) to solve for x

    x^2-(6\sqrt{3})x+24=0

    \left (x-\frac{6\sqrt{3}}{2}\right ) ^2-\frac{108}{4}+\frac{96}{4}=0

    \left ( x-\frac{6\sqrt{3}}{2}+\frac{2\sqrt{3}}{2}\right )\left ( x-\frac{6\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right )=0

    (x-2\sqrt{3})(x-4\sqrt{3})=0

    So, x=2\sqrt{3} or  x=4\sqrt{3}

    Hope that helps
    i wish it did but i don't understand any of it. i have never used that quadratic formula. oh i think you're right... that x wasn't meant to be squared. really sorry to anyone who tried to do it with the wrong equation.
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  12. #12
    Newbie Miasmagasma's Avatar
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    Quote Originally Posted by Miasmagasma View Post
    i wish it did but i don't understand any of it. i have never used that quadratic formula. oh i think you're right... that x wasn't meant to be squared. really sorry to anyone who tried to do it with the wrong equation.
    i now recognise the formula actually but i'm confused about the second step ( where the 2 brackets = 0 ) and the final step. can you walk me through like you were talking to a dimwit?
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  13. #13
    Senior Member Stroodle's Avatar
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    You should have a look here: Practical Algebra Lessons under the "intermediate algebra topics"; it explains quadratics a lot better than I'm able to.

    For the method I used (completing the square) go to Completing the Square: Solving Quadratic Equations otherwise there's also a link there on how to use the quadratic formula - which is essentially the same thing as it's derived by completing the square - where you can just plug in values into the formula to solve for x
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  14. #14
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    Yes you may use the quadratic formula as what Prove It said..
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  15. #15
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    Quote Originally Posted by Miasmagasma View Post
    solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

    i can't isolate the '6√3' without getting something wrong and ruining the equation.

    Hi all, I am a new member of forum. Would a newcomer be warmly welcome here? Good day you guys!!!

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