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Math Help - Roots of a Quadratic equation

  1. #1
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    Question Roots of a Quadratic equation

    Given that 3p(x^2) - 7qx + 3p = 0 has equal roots
    and p and q are positive,
    find the ratio p : q
    and solve the equation.

    (I was able to deduce the relation 49q^2 = 36p, but how is the ratio further on calculated?
    How are the value of p and q deduced in such a way that substituting their values would solve the equation? )

    (The answers provided are 7 : 6 and 1.)
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  2. #2
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    \Delta = (-7q)^2-4\cdot 3p\cdot 3p=49q^2-36p^2
    Because the equation has equal roots \Delta=0\Rightarrow 49q^2=36p^2\Rightarrow \frac{49}{36}=\frac{p^2}{q^2} \Rightarrow \frac{7}{6}=\frac{p}{q}

    3px^2-7qx+3p=0\Leftrightarrow 3\frac{p}{q}x^2-7x+3\frac{p}{q}=0 Solve it.
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  3. #3
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    Quote Originally Posted by Ilsa View Post
    Given that 3p(x^2) - 7qx + 3p = 0 has equal roots
    and p and q are positive,
    find the ratio p : q
    and solve the equation.

    (I was able to deduce the relation 49q^2 = 36p, but how is the ratio further on calculated?
    How are the value of p and q deduced in such a way that substituting their values would solve the equation? )

    (The answers provided are 7 : 6 and 1.)
    You have a small error. It should be 49q^2 = 36p^2 from which it follows that \frac{p^2}{q^2} = \frac{49}{36}. Can you take it from here?
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  4. #4
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    Another way to do this: if we call the two equal roots "r" then we must have
    3p(x- r)^2= 3p(x^2- 2rx+ r^2)= 3px^2- 6prx+ 3pr^2= 3px^2- 7qx+ 3p
    and so must have 6pr= 7q and 3pr^2= 3p. That second equation reduces to r^2= 1 so that r= 1 or r= -1. In order that p and q both be positive, we must have r= 1 and then we have 6p= 7q so that \frac{p}{q}= \frac{7}{6}.
    Last edited by HallsofIvy; April 2nd 2011 at 08:24 AM.
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  5. #5
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    Hello, Ilsa!

    \text{The answers provided are: }7\!:\!6 \text{ and } 1.
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . \uparrow
    . . . . . . . . . . . . . . . . . . . . . . . . . . . . ?

    Are they saying that p = q is a solution?

    This is not true . . .

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  6. #6
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    "Are they saying that p=q is a solution?" No, they are not saying that...
    \frac{7}{6} is \frac{p}{q} and 1 is the solution of that equation.
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