Results 1 to 6 of 6

Math Help - solving an equation which contains x and sqrtx

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    16

    solving an equation which contains x and sqrtx

    Techniques of Differentiation
    When 3x+4\sqrt{x}-1=0 then why is \sqrt{x}=\frac{-2+\sqrt{7}}{3}?
    It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with x wouldn't do any good since I would get x\sqrt{x}. Same if I would quadratize the equation by itself.
    I think that they used the quadratic formula somehow but I don't see how.
    Obviously the problem is the combination of
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2006
    Posts
    244
    Quote Originally Posted by Sebastian de Vries View Post
    Techniques of Differentiation
    When 3x+4\sqrt{x}-1=0 then why is \sqrt{x}=\frac{-2+\sqrt{7}}{3}?
    It wouldn't be a problem for me if the equation would be in the normal quadratic form. Multiplying all terms with x wouldn't do any good since I would get x\sqrt{x}. Same if I would quadratize the equation by itself.
    I think that they used the quadratic formula somehow but I don't see how.
    Obviously the problem is the combination of
    Put y=\sqrt{x} to get the quadratic:

    3y^2+4y-1=0

    Now use the quadratic formula and you get two solutions, one of which is negative and so is a spurious solution since y=\sqrt{x}\ge0, the remaining positive root gives the solution.

    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Hmm, when I do this I get:

    \sqrt{x}=\frac{-4\pm\sqrt{28}}{6}

    Not sure if this is something I've missed or what?

    My above answer was done using the quadratic forumula by the way
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by craig View Post
    Hmm, when I do this I get:

    \sqrt{x}=\frac{-4\pm\sqrt{28}}{6}

    Not sure if this is something I've missed or what?
    \sqrt{28} = 2 \sqrt{7} ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    how did I miss that one lol.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2008
    Posts
    16
    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find derivative of sqrtx...
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: April 19th 2010, 09:18 AM
  2. Solving equation
    Posted in the Algebra Forum
    Replies: 8
    Last Post: January 11th 2010, 01:14 PM
  3. Equation solving
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2010, 12:04 AM
  4. solving diff equation w/ bournoulli's equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 14th 2008, 06:06 PM
  5. Solving a Log Equation
    Posted in the Algebra Forum
    Replies: 8
    Last Post: September 30th 2007, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum