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Math Help - Quadratics # 2

  1. #1
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    Question Quadratics # 2

    Prove that there are always two real values of k for which ax^2 + 2bx + c + k(x^2 + 1) is a perfect square, unless a =c, and b=0.

    What I did was:

    For the equation to be a perfect square, its roots must be equal.
    For equal roots, the discriminant must be equal to 0.
    Thus, discriminant = (2b)^2 - 4(a+k)(k+c) = 0

    Eventually, I got 4b^2 - 4ak - 4k^2 - 4ac - 4kc = 0 but I don't know how to apply this to the problem.

    Please, any help?

    Thanx
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by xwrathbringerx View Post
    Eventually, I got 4b^2 - 4ak - 4k^2 - 4ac - 4kc = 0 but I don't know how to apply this to the problem.
    This equality can be written k^2+(a+c)k+ac-b^2=0... this is a quadratic equation.
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  3. #3
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    ah so the roots of this quadratic equation are the two real values that will make ax^2 + 2bx + c + k(x^2 + 1) a perfect square unless a = c and b = 0 where it is only k^2 = 0? (or have I got the wrong idea)
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    ah so the roots of this quadratic equation are the two real values that will make ax^2 + 2bx + c + k(x^2 + 1) a perfect square unless a = c and b = 0 where it is only k^2 = 0?
    Yes.
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