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Math Help - Real Roots

  1. #1
    Junior Member casey_k's Avatar
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    Post Real Roots

    show that if k is any real number, the equation x^2 + (k+1)x + k = 0 always has real roots. For what value of k are the roots equal?
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  2. #2
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    Quote Originally Posted by casey_k View Post
    show that if k is any real number, the equation x^2 + (k+1)x + k = 0 always has real roots. For what value of k are the roots equal?
    \text{descriminant} = {b^2 - 4ac}

      = {(k+1)^2 - 4(a)(k)}

      ={k^2+2k+1 - 4k}

      =  {k^2-2k+1}

    For these to be real

     k^2 - 2k + 1 \geq  0

     (k-1)^2  \geq  0

    Is this true for all real numbers ?

    For these to be EQUAL

     k^2 - 2k + 1 = 0

     (k-1)^2  = 0

     k-1  =  0

     k  =  1
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