Results 1 to 7 of 7

Math Help - solution of equation

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    184
    Thanks
    1

    solution of equation

    Consider the following equation,

    x^{2}+\sqrt{x}-2=0
    (\sqrt{x}+2)(\sqrt{x}-1)=0
    \sqrt{x}=1 or \sqrt{x}=-2
    For the equation \sqrt{x}=-2, can I say that there is "no real solution".
    Thus x=1

    My teacher mark me wrong for the part "no real solution", she wrote "wrong reason for rejection".
    Can someone please enlighten me? Am I correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by acc100jt View Post
    Consider the following equation,

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

    Mr F says: One of the above two lines is wrong because they are not consistent with each other (expand the second line out to see this).

    \sqrt{x}=1 or \sqrt{x}=-2
    For the equation \sqrt{x}=-2, can I say that there is "no real solution".
    Thus x=1

    My teacher mark me wrong for the part "no real solution", she wrote "wrong reason for rejection".
    Can someone please enlighten me? Am I correct?
    Until you resolve the above problem (in red) nothing more can be done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    184
    Thanks
    1
    sorry, my mistake. The equation should be x+\sqrt{x}-2=0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,393
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by acc100jt View Post
    sorry, my mistake. The equation should be x+\sqrt{x}-2=0
    x+\sqrt{x}-2=(\sqrt{x}+2)(\sqrt{x}-1) is correct.
    But I think that you need to ask your teacher the question.
    It seems to be her preference which we have no way of knowing.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2006
    Posts
    184
    Thanks
    1
    the fatorisation is correct. It is the part when I say "no real solution" that she marked me wrong
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,393
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by acc100jt View Post
    the fatorisation is correct. It is the part when I say "no real solution" that she marked me wrong
    I understand the OP.
    But none of us here is likely to be your teacher.
    That means we do not know why she did not like you answer.
    Therefore, you must ask her.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,974
    Thanks
    1121
    When you said "no real solution" did you make it clear that you were only talking about the " \sqrt{x}= -2" term? Because x= 1 is clearly as solution and you get that from the other factor, \sqrt{x}- 1= 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Equation solution
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 1st 2010, 06:53 AM
  2. Replies: 2
    Last Post: May 18th 2009, 12:51 PM
  3. Solution to equation
    Posted in the Advanced Math Topics Forum
    Replies: 7
    Last Post: April 26th 2008, 08:51 PM
  4. Solution to this equation
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 16th 2007, 01:02 PM
  5. Solution of equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: June 19th 2006, 07:52 AM

Search Tags


/mathhelpforum @mathhelpforum