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I need help with an inequality question

  1. Oct 2, 2004 #1
    what are the range of values of k that gives the equation (k+1)x^2+4kx+9=0 ...I work it out :confused: ...please help
     
  2. jcsd
  3. Oct 2, 2004 #2

    arildno

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    Welcome to PF!
    I think you have omitted something from your text; what was that?
     
  4. Oct 2, 2004 #3
    ups sorry...it should say at the end...that gives the equation no real roots
     
  5. Oct 2, 2004 #4

    arildno

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    And what do you think that means?
     
  6. Oct 2, 2004 #5

    shmoe

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    Hi, you can use the quadratic formula to find the roots of your equation. Forcing no real roots is equivalent to forcing the discriminant (the part under the root sign) to be negative. This will give you conditions on k that you're looking for.

    ps. you have to make a restriction on k to guarantee that your equation is a quadratic and not linear. what is this restriction?
     
  7. Oct 2, 2004 #6
    i cant use the quadrativ formula for this euqation..when i work out the bracket i get
    x^2+x^2k+4kx+9=0 ...i have no idea how to use the discrimant in this case..please help
     
  8. Oct 2, 2004 #7

    arildno

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    Let a=k+1, b=4k, c=9.
    Then your equation looks like
    ax^2+bx+c=0
    Can you solve that one?
     
  9. Oct 2, 2004 #8
    i know that the discrimiant has to be smaller than 0....
     
  10. Oct 2, 2004 #9

    arildno

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    And what is the discriminant, expressed with a,b and c?
     
  11. Oct 2, 2004 #10
    no i am still confused
     
  12. Oct 2, 2004 #11

    arildno

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    What is your problem?
     
  13. Oct 2, 2004 #12
    could you please just show me how i work out this questions?
     
  14. Oct 2, 2004 #13
    because then i will undertand
     
  15. Oct 2, 2004 #14

    arildno

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    I ask you again:
    Given the equation:
    ax^2+bx+c=0
    What is the discriminant?
     
  16. Oct 2, 2004 #15
    16k^2-(4[k+1]*9)<0 that becomes 16k^2-36k-36<0
     
  17. Oct 2, 2004 #16

    shmoe

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    Good, now can you find the values of k that satisfy this new inequality?
     
  18. Oct 2, 2004 #17

    arildno

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    Very good!
    Here's a hint:
    In order to find the range of k-values your after,
    1. find the zeros in your discriminant.
    That is, solve the equation for k:
    [tex]16k^{2}-36k-36=0[/tex]
    2. You weren't interested in the k-values for which the discriminant was zero, but the k-values for which the discriminant is less than zero.
    But you should figure out for yourself that those values must lie between the two values found in 1.
     
  19. Oct 2, 2004 #18
    i see that i have to factoris ethe equation in the disriminant now..but i cant find the right numbers
     
  20. Oct 2, 2004 #19

    arildno

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    Quite true!
    Look at my previous post for hints.
     
  21. Oct 2, 2004 #20
    please...i have no idea how to continue...
     
  22. Oct 2, 2004 #21

    arildno

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    Well, what values of k solves:
    [tex]16k^{2}-36k-36=0[/tex] ?

    (Note: You were asked to find the values of k so that the discriminant is less than zero, not zero, but finding the zeroes is a good start)
     
  23. Oct 2, 2004 #22
    k..thanks for your help...
     
  24. Oct 2, 2004 #23

    arildno

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    Now, having found the k-values yielding zero discriminant, you should be able to write the discrimanant as:
    [tex]16k^{2}-36k-36=16(k-3)(k+\frac{3}{4})[/tex]
    What must then the k-interval be which yields negative discriminant?
     
  25. Oct 2, 2004 #24
    how did u factorise that?
     
  26. Oct 2, 2004 #25

    arildno

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    It is the roots of equation gained by setting the discriminant equal to 0 (that is, -3/4 and 3 are the roots)
     
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