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Given (2xsquared +1 )/xsquared minus 2x=k find the values of k for which real roots exist?

plz include the steps btw this question is worth 10 marks for sum reason

2 Answers

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  • Aleko
    Lv 6
    8 years ago
    Favorite Answer

    (2x^2+1)/(x^2-2x)=k 2x^2+1=k(x^2-2x)

    x^2(2-k)+2kx+1=0

    for real roots the discriminant must be>0 so 4k^2-8+4k>0 (k+2)(k-1)>0

    k>-2 and k>1

    so there are real roots for values of k>1

  • 8 years ago

    (2x^2 + 1)/(x^2 - 2x) = k

    2x^2 + 1 = kx^2 - 2kx

    (2 - k)x^2 + 2kx + 1 = 0

    Real roots exist when the determinant ≥ 0.

    (2k)^2 - 4(2 - k)(1) ≥ 0

    4k^2 - 8 + 4k ≥ 0

    k^2 + k - 2 ≥ 0

    (k + 2)(k - 1) ≥ 0

    k ≥ 1 or k ≤ 2

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