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Deepak Bora
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Quadratic equations, find the value of k for which the roots are real and equal

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Find the value of k for which the equation x2 + k(2x + k – 1) + 2 = 0 has real and equal roots

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  1. x²+ k(2x+k-1)+2=0

    x²+2kx+k²-k+2=0

    After comparing ax²+bx+c=0

    a=1, b= 2k & c= k²-k+2

    Because our roots are real and equal

    ∴ D= 0

    b²-4ac= 0

    (2k)²-4(1)(k²-k+2)=0

    4k²-4k²+4k-8=0

    4k=8

    ∴ K= 2

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