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Quadratic Roots

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Barry has just solved a quadratic equation. He sees that the roots are rational, real, and unequal. This means the discriminant is

    a) zero, b) negative, c) a perfect square, d) a non perfect square

    2. Relevant equations

    3. The attempt at a solution

    I think the answer is d) a non perfect square

    if the roots are real and rational then the discriminant cant be negative, and if they are unequal then the discriminant cant be a perfect square

    is this the right way to do this problem?
  2. jcsd
  3. Mar 8, 2008 #2

    Why do you think so?? You are expected to show your work before anyone here can help you!! Ok, the general form of the quadratic eq is:

    [tex]ax^{2}+bx+c=0[/tex] the formula for the discriminant is

    [tex] D=b^{2}-4ac[/tex] right?

    The formula for the two roots is:

    [tex] x_1,_2=\frac{-b+-\sqrt D}{2a}[/tex], so you want your answer to be a rational nr, and the roots to be distinct, right?
    This means:

    [tex] x_1=\frac{-b-\sqrt D}{2a}= \frac{m}{n} \ (not \ equal \ to)=/=x_2=\frac{-b+\sqrt D}{2a}=\frac{p}{q}[/tex] where m,n,p,q are integers.

    So what do you think now????????? What would happen if, say D=3, D=4, D=0, or D<0??
    Last edited: Mar 8, 2008
  4. Mar 8, 2008 #3


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    Not exactly. Look at the general solution to a quadratic equation of one variable. What kind of solution occurs when the discriminant fits each of the choices in your question? What kind of discriminant will give you TWO solutions which are rational and real and unequal?
  5. Mar 10, 2008 #4
    I see. the discriminant should be a perfect square. thanks.
  6. Mar 10, 2008 #5
    You're welcome! Just make sure to show some work of yours next time!!
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