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Math Help - Distinct Roots

  1. #1
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    Distinct Roots

    Find the values of k when the equation kx^2+k=8x-2xk has two real distinct roots?

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     kx^2+k=8x-2xk

     kx^2+k-8x+2xk=0

     kx^2-8x+2xk+k=0

     kx^2+(-8+2k)x+k=0

    A quadratic equation ax^2 +bx+c=0 will have two distinct real roots when b^2-4ac>0

    In your case solve (-8+2k)^2-a\times k\times k>0
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    Thanks !
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    Quote Originally Posted by p4pri View Post
    Find the values of k when the equation kx^2+k=8x-2xk has two real distinct roots?
    Write the equation as kx^2+(2k-8)x+k.
    Using the discriminate we want (2k-8)^2-4(k)(k)>0.
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  5. #5
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    Thx for ur ans.....I cudnt find d way of thinking d ans 4 dis qn.
    Suppose for fixed real numbers a and b, f suffix c(x)=x cube+ax square+bx+c has 3 distinct roots for c=0.Then
    (1)f suffix c(x) has 3 distinct roots for all real c
    (2)f suffix c(x) has 3 distinct roots for all real c>0 or for all real c<0, but not for both
    (3)f suffix c(x) has 3 distinct roots for all real c in (p,q) for some p<0 and q>0
    (4)f suffix c(x)need not have 3 distinct roots for all real c other than 0
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  6. #6
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    kx^2+k=8x-2xk,

    Discriminant = D = (2k - 8)^2 - 4(k)(k) > 0, two real distict roots

    k = -4, D = 256 - 64 = 192

    k = - 3, D = 196 - 36 = 160

    k = -2, D = 144 - 16 = 128

    k = -1, D = 100 - 4 = 96

    k = 0, D = 64 - 0 = 64

    k = 1, D = 36 - 4 = 32

    k = 2, D = 0,

    k = 3, D = 4 - 36 = -34

    k = 4, D = 0 - 64 = - 64

    k = 5 D = 4 - 100 = - 96

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