Results 1 to 11 of 11

Math Help - Nature of Roots

  1. #1
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17

    Nature of Roots

    Hey sup got stuck with nature of roots, so sad -_-

    -> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

    and

    -> Find the values of k for which the roots of the equation
    kx^2 + 2(k-3)x + k - 5 = 0 are real. Give 3 values for k for which the roots are rational.

    Thanx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    1) mx^2+5x-5m=0

    Use the quadratic formula and both claims will be demonstrated.

    2) Look at the equation like this: kx^2+(2k-6)x+(k-5)

    The discriminant or, \sqrt{b^2-4ac} part, determines the real/imaginary set of the roots. So in order for the whole root to be real (assuming coefficients are real as well), \sqrt{(2k-6)^2-4(k)(k-5)} must be real, meaning that (2k-6)^2-4(k)(k-5)>0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Quote Originally Posted by NeF
    Hey sup got stuck with nature of roots, so sad -_-

    -> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m
    Try using the discriminant.

    b^{2}-4ac

    25-4(m)(-5m)

    The roots are real and unequal if the discriminant is positive.

    25+20m^{2} is positive for all real values of m.
    Last edited by CaptainBlack; July 8th 2006 at 11:02 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17
    [QUOTE=galactus][QUOTE=NeF]Hey sup got stuck with nature of roots, so sad -_-

    -> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

    Try using the discriminant.

    b^{2}-4ac

    25-4(m)(-5m)

    The roots are real and unequal if the discriminant is positive.

    25+20m^{2} is positive for all real values of m.
    I got as far as 25+20m^{2} would the next step be 5+4m^{2} ?
    Last edited by NeF; July 8th 2006 at 02:57 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You can do that if you like, but it's done. The square assures you that all real values of m are positive.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by NeF
    Quote Originally Posted by galactus
    Quote Originally Posted by NeF
    Hey sup got stuck with nature of roots, so sad -_-
    -> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

    I got as far as 25+20m^{2} would the next step be 5+4m^{2} ?
    Actually, no, becuase 25+20m^2\neq5+4m^2 however, it doesn't actually matter becuase you've already shown that the discriminant is positive.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,937
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Jameson
    1) mx^2+5x-5m=0

    Use the quadratic formula and both claims will be demonstrated.
    Though technically we have to exclude m = 0 from this statement...

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17

    Question

    Thanks for your help guys. One thing I realised is that I should post how far I got with each sum when posting questions.

    This is as far as I got with the second* question. Please let me know if I am doing something wrong cause all I did was random stuff x/

    kx^{2} + 2(k-3)x + k - 5 = 0<br /> <br />
  \sqrt{b^2-4ac}<br /> <br />
 (2k-6)^2-4(k)(k-5)>0<br /> <br />
 (2k-6)(2k-6) - 4(k^2-5k) > 0<br /> <br />
  4k^2-12k-12k+36-4k^2+20k>0<br /> <br />
  -4k = -36<br />
  divide by -4<br /> <br />
    k = 9

    <br />
kx^2 + 2(k-3)x + k - 5 = 0
      9x^2+2(9-3)x+9-5 = 0

    9x^2+12x+4=0<br /> <br />
(3x+2)(3x+2)=0
    (3x+2)^2 = 0

    Then take (3x+2) = 0
      x = -2/3
    and Sub the x and k values into
    <br />
kx^2 + 2(k-3)x + k - 5 = 0
     (9)(-2/3)^2+2(9-3)-2/3+9-5=0

    So all 3 rational values are  9 ; (3x+2)^2 ; 0
    Is that correct and btw the sum is worth 12 marks.
    Sorry if I come across as a complete doos
    Last edited by NeF; July 9th 2006 at 01:21 PM. Reason: I mean 2nd Question of mine
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by NeF
    Thanks for your help guys. One thing I realised is that I should post how far I got with each sum when posting questions.

    This is as far as I got with the second* question. Please let me know if I am doing something wrong cause all I did was random stuff x/

    kx^{2} + 2(k-3)x + k - 5 = 0<br /> <br />
  \sqrt{b^2-4ac}<br /> <br />
 (2k-6)^2-4(k)(k-5)>0<br /> <br />
 (2k-6)(2k-6) - 4(k^2-5k) > 0<br /> <br />
  4k^2-12k-12k+36-4k^2+20k>0<br /> <br />
  -4k = -36<br />
  divide by -4<br /> <br />
    k = 9

    <br />
kx^2 + 2(k-3)x + k - 5 = 0
      9x^2+2(9-3)x+9-5 = 0

    9x^2+12x+4=0<br /> <br />
(3x+2)(3x+2)=0
    (3x+2)^2 = 0

    Then take (3x+2) = 0
      x = -2/3
    and Sub the x and k values into
    <br />
kx^2 + 2(k-3)x + k - 5 = 0
     (9)(-2/3)^2+2(9-3)-2/3+9-5=0

    So all 3 rational values are  9 ; (3x+2)^2 ; 0
    Is that correct and btw the sum is worth 12 marks.
    Sorry if I come across as a complete doos
    Discriminant is 36 - 4k
    Roots are real for k\leq9
    For rational roots 36 - 4k is a perfect square.

    Keep Smiling
    Malay
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    This is an incorrect statement...
    Quote Originally Posted by NeF
    <br /> <br />
  4k^2-12k-12k+36-4k^2+20k>0<br /> <br />
  -4k = -36<br />
  divide by -4<br /> <br />
    k = 9
    There are two things wrong with your work, you can't change > into = whenever you feel like it, and it should say >= 0

    here is my work...

    4k^2-12k-12k+36-4k^2+20k\geq0

    -4k+36\geq0

    -4k\geq-36

    k\leq9 note:when multiplying or dividing by a negative number, you need to flip the signs.
    Last edited by Quick; July 10th 2006 at 04:38 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    NeF
    NeF is offline
    Newbie
    Joined
    Jul 2006
    From
    Cape Town, RSA
    Posts
    17

    Thumbs up

    Awesome thanks for pointing out my mistakes
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. nature of roots
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 15th 2012, 07:32 AM
  2. about nature of roots
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 2nd 2012, 05:04 PM
  3. Nature of a point
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 3rd 2010, 05:47 AM
  4. Quadratic Equation - Nature of roots
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: January 17th 2010, 11:50 PM
  5. nature of roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 18th 2009, 07:12 AM

Search Tags


/mathhelpforum @mathhelpforum