Results 1 to 3 of 3

Thread: Roots Independent of lambda.

  1. #1
    Member
    Joined
    Dec 2010
    Posts
    106
    Thanks
    2

    Roots Independent of lambda.

    If $\displaystyle x^5+5\lambda x^4+(\lambda a-4)x^2-(8\lambda+3)x+\lambda a-2=0$. Then the value of $\displaystyle a$ for which the Given equation has one root Independent of $\displaystyle \lambda$

    Here answer are
    (i) $\displaystyle \frac{64}{5}$

    (ii)$\displaystyle -\frac{64}{5}$

    (iii) $\displaystyle 3$

    (iv)$\displaystyle -3$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    Pick a value for lambda (1,2,3,...). Since this is multiple choice, you should be able to eliminate the wrong choices (plot the graphs).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    Quote Originally Posted by jacks View Post
    $\displaystyle x^5+5\lambda x^4+(\lambda a-4)x^2-(8\lambda+3)x+\lambda a-2=0$. Then the value of $\displaystyle a$ for which the Given equation has one root Independent of $\displaystyle \lambda$
    In general, for a family of polynomials:

    $\displaystyle \{f_{\lambda,\;a}(x)=p(x)+\lambda q_a(x)\}$

    if $\displaystyle r$ is a root of $\displaystyle p(x)$ and $\displaystyle q_a(x)$ then, $\displaystyle r$ is a root of $\displaystyle f_{\lambda,\;a}(x)$. So, finding a root $\displaystyle r$ of $\displaystyle p(x)$ and substituting: $\displaystyle q_a(r)=0$ , we obtain a sufficient condition for $\displaystyle a$.

    That is a pity, if $\displaystyle r=2$ were a root of $\displaystyle p(x)$ (it isn't) then,

    $\displaystyle q_a(2)=0 \Leftrightarrow a=-64/5$

    Possiibly (of course I'm not sure) there is typo.

    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Eigenvalues of y''+2y'+\lambda y=0
    Posted in the Differential Equations Forum
    Replies: 11
    Last Post: Feb 26th 2011, 05:46 PM
  2. Replies: 1
    Last Post: Oct 29th 2009, 04:58 PM
  3. Lambda calculus
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 13th 2009, 09:02 PM
  4. Replies: 5
    Last Post: Jan 18th 2009, 03:39 AM
  5. show that V(X) = lambda
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Jan 11th 2009, 03:50 AM

Search Tags


/mathhelpforum @mathhelpforum