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Thread: 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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    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?
    $\displaystyle \text{descriminant} = {b^2 - 4ac} $

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

    $\displaystyle ={k^2+2k+1 - 4k}$

    $\displaystyle = {k^2-2k+1} $

    For these to be real

    $\displaystyle k^2 - 2k + 1 \geq 0 $

    $\displaystyle (k-1)^2 \geq 0 $

    Is this true for all real numbers ?

    For these to be EQUAL

    $\displaystyle k^2 - 2k + 1 = 0 $

    $\displaystyle (k-1)^2 = 0 $

    $\displaystyle k-1 = 0 $

    $\displaystyle k = 1 $
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