Page 2 of 2 FirstFirst 12
Results 16 to 30 of 30

Math Help - Solving Quadratic Systems

  1. #16
    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 BlueStar View Post
    You mean y = 9 +- 119i / -2?
    no, y = \frac {9 \mp \sqrt {119}~i}{2}
    Last edited by Jhevon; July 23rd 2007 at 05:50 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by BlueStar View Post
    You mean y = 9 +- 119i / -2?
    Actually, no.

    y = -x^2 = - \left ( \frac{-9 \pm i \sqrt{119}}{2} \right ) = \frac{9 \mp i \sqrt{119}}{2}

    (There was a typo about the square root sign earlier.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #18
    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 topsquark View Post

    (There was a typo about the square root sign earlier.)
    yes, you are correct. i shall change my posts
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Okay.


    I'll have to be led through on solving this. I cannot brain today, I have the dumb.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    x^4 + 9x^2 + 50 = 0
    The quadratic formula says that
    x^2 = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 50}}{2 \cdot 1} = \frac{-9 \pm \sqrt{-119}}{2}

    So
    x^2 = \frac{-9 \pm i \sqrt{119}}{2}

    A brief diversion:
    y = -x^2 = - \left ( \frac{-9 \pm i \sqrt{119}}{2} \right ) = \frac{9 \mp i \sqrt{119}}{2}

    Again,
    x^2 = \frac{-9 \pm i \sqrt{119}}{2}

    So
    x =\pm \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} <-- The two \pm symbols do not have any relation to each other.

    This is technically an unsimplified answer, but in your case I'd leave it like this.

    So the solution set is the four points:
    The two points
    \left ( \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )
    and the two points
    \left (- \sqrt{\frac{-9 \pm i \sqrt{119}}{2}} , \frac{9 \mp i \sqrt{119}}{2} \right )

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Are those supposed to be the final solutions? Because that seems wrong. It's for a multiple question, and the solution answers only consist of variations of 0 and 5.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    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 BlueStar View Post
    Are those supposed to be the final solutions? Because that seems wrong. It's for a multiple question, and the solution answers only consist of variations of 0 and 5.
    0 and 5 are not solutions to the equations you've posted. the minus sign in from of (y - 5)^2 did away with that possibility. 0 would be a solution for x if it was plus
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Quote Originally Posted by Jhevon View Post
    x = 0 is NOT a solution (if we have x = 0, that would lead to -25 = 25), and if it was, since y = -x^2, if x = 0, then y = 0, so a solution would be (0,0)
    What about this? That's the only possible solution that's on the multiple choice.


    I don't think I wrote the question wrong.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    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 BlueStar View Post
    What about this? That's the only possible solution that's on the multiple choice.


    I don't think I wrote the question wrong.
    i told you, (0,0) is only a solution if you have a + (y - 5)^2. otherwise, there are NO real solutions. if you did not type the question wrong, topsquark's solutions are correct and the solutions you have are wrong. go back to the question and make sure you have the signs right
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Nope, the signs are right.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    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 BlueStar View Post
    Nope, the signs are right.
    then topsquark's solution is correct
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Then either the quiz is wrong or topsquark is, or something weird's going on.
    Follow Math Help Forum on Facebook and Google+

  13. #28
    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 BlueStar View Post
    Then either the quiz is wrong or topsquark is, or something weird's going on.
    something weird is going on
    Follow Math Help Forum on Facebook and Google+

  14. #29
    Junior Member
    Joined
    Jul 2007
    Posts
    49
    Well, I just selected the answer as (0,0) and it was correct.
    Follow Math Help Forum on Facebook and Google+

  15. #30
    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 BlueStar View Post
    Well, I just selected the answer as (0,0) and it was correct.
    that simply means the minus should have been a plus. good going. it was the best choice of all that were there
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. linear quadratic systems
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 29th 2010, 06:46 AM
  2. Modelling a quadratic - Systems of linear equations
    Posted in the Pre-Calculus Forum
    Replies: 30
    Last Post: May 12th 2010, 06:13 AM
  3. Solving Quadratic Systems of Equations
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 10th 2008, 03:11 PM
  4. Rotation and Systems of Quadratic Equations
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: June 4th 2007, 08:26 PM
  5. Replies: 3
    Last Post: October 11th 2006, 10:15 PM

Search Tags


/mathhelpforum @mathhelpforum