Results 1 to 13 of 13

Math Help - Solving for x

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    11

    Solving for x

    Solve for x: x^2+1=x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by DCHomage View Post
    Solve for x: x^2+1=x
    this is a quadratic equation, do you know how to deal with those?

    first note that you have x^2 - x + 1 = 0

    next you would try to factor, but it won't work here, so the next move is to complete the square, or use the quadratic formula. can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    11
    Let's see. Do I move the -x+1 over?
    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 DCHomage View Post
    Let's see. Do I move the -x+1 over?
    Quote Originally Posted by Jhevon View Post
    this is a quadratic equation, do you know how to deal with those?

    first note that you have x^2 - x + 1 = 0

    next you would try to factor, but it won't work here, so the next move is to complete the square, or use the quadratic formula. can you continue?
    Do you know what completing the square means? Are you familiar with the quadratic formula?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by DCHomage View Post
    Let's see. Do I move the -x+1 over?
    were that the case, i would have done it in the first step.

    the next move, i said, is to complete the square, or use the quadratic formula. are you familiar with these methods?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2009
    Posts
    11
    can't remember. I have been working at my full time job that i forgot about it.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by DCHomage View Post
    can't remember. I have been working at my full time job that i forgot about it.
    the easier way would be the quadratic formula. it says the following

    the solution(s) to an equation of the form ax^2 + bx + c = 0 are given by x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

    what would that give as the answer to this problem?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2009
    Posts
    11
    Not sure. I have this so far
    x = \frac {x \pm \sqrt{x^2 - 4x^2}}{2x^2}<br />
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DCHomage View Post
    Not sure. I have this so far
    x = \frac {x \pm \sqrt{x^2 - 4x^2}}{2x^2}
    No.

    Compare x^2 - x + 1 with ax^2 + bx + c and it should be clear that a = 1, b = -1 and c = 1. These are the values that get substituted into the quadratic formula.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2009
    Posts
    11
    I got this so far.

     x = \frac {1 \pm \sqrt{-3}}{2}
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DCHomage View Post
    I got this so far.

     x = \frac {1 \pm \sqrt{-3}}{2}
    Yes. So there's no real solution. (There are however two non-real solutions).
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Jan 2009
    Posts
    11
    So, I can't solve for x?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by DCHomage View Post
    So, I can't solve for x?
    Quote Originally Posted by mr fantastic View Post
    [snip]
    there's no real solution. (There are however two non-real solutions).
    ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solving for x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 7th 2009, 05:13 PM
  2. help solving for y
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 1st 2009, 11:25 AM
  3. help me in solving this
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 13th 2008, 02:51 PM
  4. Solving x'Px = v
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 11th 2008, 04:21 PM
  5. Replies: 3
    Last Post: October 11th 2006, 10:15 PM

Search Tags


/mathhelpforum @mathhelpforum