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Thread: Roots of a Quadratic equation

  1. #1
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    Question Roots of a Quadratic equation

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

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

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

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

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

    (The answers provided are $\displaystyle 7 : 6 $ and $\displaystyle 1$.)
    You have a small error. It should be $\displaystyle 49q^2 = 36p^2$ from which it follows that $\displaystyle \frac{p^2}{q^2} = \frac{49}{36}$. 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
    $\displaystyle 3p(x- r)^2= 3p(x^2- 2rx+ r^2)= 3px^2- 6prx+ 3pr^2= 3px^2- 7qx+ 3p$
    and so must have $\displaystyle 6pr= 7q$ and $\displaystyle 3pr^2= 3p$. That second equation reduces to $\displaystyle r^2= 1$ 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 $\displaystyle 6p= 7q$ so that $\displaystyle \frac{p}{q}= \frac{7}{6}$.
    Last edited by HallsofIvy; Apr 2nd 2011 at 08:24 AM.
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  5. #5
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    Hello, Ilsa!

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

    Are they saying that $\displaystyle p = q$ 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...
    $\displaystyle \frac{7}{6}$ is $\displaystyle \frac{p}{q}$ and 1 is the solution of that equation.
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