Results 1 to 8 of 8

Math Help - Showing that the equation has real roots for all real values of 'k'

  1. #1
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54

    Question Showing that the equation has real roots for all real values of 'k'

    Hello everyone here on Math Help Forum.

    I have an urgent question here: show that the equation x^2 + kx = 4 - 2k has real roots for all real values of k.

    What I know is that the term ''real roots'' imply that b^2 - 4ac is equal or more than 0. But my primal problem is that I have difficulties in showing and proving. So here are my workings:

    x^2 + kx = 4 - 2k
    x^2 + kx -4 +2k = 0

    Here a = 1, b = k, c = -4 + 2k

    b^2 -4ac = k^2 - 4(1)(-4+2k)
    k^2 -8k +24
    (x-4)(x-4)
    (x-4)^2

    I have tried my best. Please pinpoint my mistakes and kindly continue the rest of the question. Thank you so much for your help and have a nice day!
    Last edited by PythagorasNeophyte; February 9th 2011 at 10:49 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    x^2 + kx - 4 + 2k =0

    a=1, b=k, c= - 4 + 2k

    Now solve for the discriminant >0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by PythagorasNeophyte View Post
    Hello everyone here on Math Help Forum.

    I have an urgent question here: show that the equation x^2 + kx = 4 - 2k has real roots for all real values of k.

    What I know is that the term ''real roots'' imply that b^2 - 4ac is equal or more than 0. But my primal problem is that I have difficulties in showing and proving. So here are my workings:

    x^2 + kx = 4 - 2k
    x^2 + kx -4 +2k = 0

    Here a = 1, b = k, c = -4 + 2k

    b^2 -4ac = k^2 - 4(1)(-4+2k)
    k^2 -8k +24
    (x-4)(x-4)
    (x-4)^2

    I have tried my best. Please pinpoint my mistakes and continue the rest of the question. Thank you so much for your help and have a nice day!
    1. k^2-8k+24\ne (x-4)^2

    2. According to the question you are asked to show that

    k^2-8k+24\geq 0 ~for\  all ~ k \in \mathbb{R}

    k^2-8k+16+8\geq 0 ~for\  all ~ k \in \mathbb{R}

    (k-4)^2+8\geq 0 ~for\  all ~ k \in \mathbb{R}

    3. A square is allways greater or equal zero. If you add a positive number to a square the sum must be greater than zero.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    Thank you so much for your replies! I am very sorry for the mistake when I should have written k instead of x when I factorise.

    But why k^2-8k+24, when factorised, gives (k-4)^2 + 8? If I were to factorise, it will be (k-4)(k-4); expanding (k-4)(k-4) gives k^2-8k+24.

    The most important thing here is to prove that x^2+kx = 4 -2k has real roots for all values of k.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by PythagorasNeophyte View Post
    Thank you so much for your replies! I am very sorry for the mistake when I should have written k instead of x when I factorise.

    But why k^2-8k+24, when factorised, gives (k-4)^2 + 8? If I were to factorise, it will be (k-4)(k-4); expanding (k-4)(k-4) gives k^2-8k+24.
    Never!
    (k-4)(k-4) = k^2-4k-4k+(-4)(-4)

    And I assure you 4 \cdot 4 \ne 24
    The most important thing here is to prove that x^2+kx = 4 -2k has real roots for all values of k.
    If the discriminant - that's what you have calculated - is greater or equal to zero then there exist real roots. Compare your own text at the beginning.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member PythagorasNeophyte's Avatar
    Joined
    Jun 2010
    Posts
    54
    Oh yes, I have forgotten that I am solving an inequality. I can only factorise in that way when I am solving an equation. I am very sorry for that.

    k^2 -8k +24 \geq 0

    \Rightarrow (k-4)^2 + 8 \geq 0

    What do I do with the 8?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by pickslides View Post
    x^2 + kx - 4 + 2k =0

    a=1, b=k, c= - 4 + 2k

    Now solve for the discriminant >0
    Using what I have already said before hand

    \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0

    \displaystyle  k^2 -4(1)(-4+2k)>0

    \displaystyle  k^2 -4(-4+2k)>0

    \displaystyle  k^2 +16-8k>0

    \displaystyle  (k-4)^2>0

    \dots
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by pickslides View Post
    Using what I have already said before hand

    \displaystyle b^2-4ac > 0 \implies k^2 -4(1)(-4+2k)>0

    \displaystyle  k^2 -4(1)(-4+2k)>0

    \displaystyle  k^2 -4(-4+2k)>0

    \displaystyle  k^2 +16-8k>0

    \displaystyle  (k-4)^2>0

    \dots
    You are using strict inequalities when you should be using $$\ge (when the discriminant is zero the original polynomial has a real double root). The discriminant has a double root at $$k=4 and since the leading coefficient is positive the discriminant is always non-negative.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove equation is -ve for all real values
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 9th 2011, 07:08 PM
  2. Real and non-real roots to a quintic equation
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: February 14th 2011, 08:42 PM
  3. A quadratic equation and its real roots
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 5th 2010, 02:57 PM
  4. Replies: 3
    Last Post: April 12th 2010, 11:37 PM
  5. Replies: 8
    Last Post: April 7th 2009, 12:15 PM

Search Tags


/mathhelpforum @mathhelpforum