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Thread: The roots of the equation

  1. #1
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    Post The roots of the equation

    There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx $ where $\displaystyle k $ is a real constant, are denoted by $\displaystyle \alpha $ and $\displaystyle \beta$.
    (a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
    (b) find the set of values of $\displaystyle k $ for which $\displaystyle \alpha $ and $\displaystyle \beta $ are real and positive. Can anyone help me with this problem?
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  2. #2
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    Quote Originally Posted by scrible View Post
    There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx $ where $\displaystyle k $ is a real constant, are denoted by $\displaystyle \alpha $ and $\displaystyle \beta$.
    (a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
    (b) find the set of values of $\displaystyle k $ for which $\displaystyle \alpha $ and $\displaystyle \beta $ are real and positive. Can anyone help me with this problem?
    HI

    $\displaystyle 9x^2+6x+1-4kx=0$

    $\displaystyle x^2+[\frac{6-4k}{9}]x+\frac{1}{9}=0$

    $\displaystyle \alpha+\beta=\frac{4k-6}{9}$

    $\displaystyle \alpha\beta=\frac{1}{9}$

    $\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\bet a}{\alpha\beta}$

    $\displaystyle =4k-6$

    $\displaystyle \frac{1}{\alpha}(\frac{1}{\beta})=9$

    the new equation would be $\displaystyle x^2-(4k-6)x+9=0 $

    then expand it

    for (b) you should realise that if the roots are real , then

    $\displaystyle b^2-4ac\geq 0$
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  3. #3
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    Quote Originally Posted by scrible View Post
    There a another problem that is hard to figure. The question ask that the roots of the equation $\displaystyle 9x^2+6x+1 = 4kx $ where $\displaystyle k $ is a real constant, are denoted by $\displaystyle \alpha $ and $\displaystyle \beta$.
    (a) Show that the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$ is $\displaystyle x^2+6x+9=4kx$
    (b) find the set of values of $\displaystyle k $ for which $\displaystyle \alpha $ and $\displaystyle \beta $ are real and positive. Can anyone help me with this problem?
    $\displaystyle 9x^2+6x+1 = 4kx $
    $\displaystyle 9x^2+6x-4kx+1 =0 $
    $\displaystyle 9x^2+(6-4k)x+1 =0 $
    if $\displaystyle ax^2+bx+c=0$
    find a, b and c from the equation...
    now, you have to determine sum and product root from the equation.
    $\displaystyle \alpha $+$\displaystyle \beta$=\frac{-b}{a}=....
    and
    $\displaystyle \alpha $ x $\displaystyle \beta$=\frac{c}{a}=....

    a) construct the equation whose roots are $\displaystyle \frac {1}{\alpha}$ and $\displaystyle \frac {1}{\beta}$
    first you have to find:
    - sum roots: $\displaystyle \frac {1}{\alpha}$+[tex]\frac {1}{\beta}=...
    - product roots:$\displaystyle \frac {1}{\alpha}$[tex]\frac {1}{\beta}=...

    after you determine the sum and product, the equation would be:
    $\displaystyle x^2-(sum)x+(product)=0$

    ....hope it help....
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  4. #4
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    Quote Originally Posted by mathaddict View Post
    HI

    $\displaystyle 9x^2+6x+1-4kx=0$

    $\displaystyle x^2+[\frac{6-4k}{9}]x+\frac{1}{9}=0$

    $\displaystyle \alpha+\beta=\frac{4k-6}{9}$

    $\displaystyle \alpha\beta=\frac{1}{9}$

    $\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\bet a}{\alpha\beta}$

    $\displaystyle =4k-6$

    $\displaystyle \frac{1}{\alpha}(\frac{1}{\beta})=9$

    the new equation would be $\displaystyle x^2-(4k-6)x+9=0 $

    then expand it

    for (b) you should realise that if the roots are real , then

    $\displaystyle b^2-4ac\geq 0$
    Thank you very much. You were of great help. I am greatful.
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