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Homework Help: Couple of algebra questions I'm struggling with

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data
    1.)
    a.) If the roots of equation px2 - 4x + 1 are equal then find value of p
    b.) Prove that the equation (k - 2)x2 + 2x - k = 0 has real roots (upside down A symbol) k E R




    2. Relevant equations



    3. The attempt at a solution
    I just don't know how to approach these questions. For a.) it says roots are equal meaning theres only 1 root but how do I find out what that root is if I have a constant in there?

    For b.) I have the same problem. I would usually plug an equation into the quadratic formula to find out if it has real roots but in this case theres a constant so I'm lost. Do I have to find this constant first and if so how?
     
  2. jcsd
  3. Jan 10, 2009 #2
    I'm pretty sure you just need to consider the discriminant.

    For a) What does the discriminant have to be for the two roots to be equal (hint: look at the quadratic equation)

    For b) Compute the discriminant
     
  4. Jan 10, 2009 #3
    Ah yea I didn't think about the fact it has to come to 0 thanks a lot. Can you verify my answer real quick I got P = 4 for a.)

    For b.) do I need to find out what k is first?
     
    Last edited: Jan 10, 2009
  5. Jan 10, 2009 #4

    Dick

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    Science Advisor
    Homework Helper

    Yes, p=4. For the second one look at the discriminant again. You can't find k. You just need to show that the discriminant is nonnegative for all k.
     
  6. Jan 10, 2009 #5
    Yes, p = 4 is correct. You should verify it by factoring the the expression and checking to see that there is only one root.

    For b), no, the question suggests that k is any real number. The idea is to show that no matter what k is, the roots of the resulting quadratic are real.

    Remember, the discriminant is the quantity under the square root. What should the sign of the discriminant be?
     
  7. Jan 10, 2009 #6
    Ah right that clears it up thanks a lot for the help.
     
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