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Math Help - 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 9x^2+6x+1 = 4kx where k is a real constant, are denoted by \alpha and \beta.
    (a) Show that the equation whose roots are \frac {1}{\alpha} and \frac {1}{\beta} is x^2+6x+9=4kx
    (b) find the set of values of k for which \alpha and \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 9x^2+6x+1 = 4kx where k is a real constant, are denoted by \alpha and \beta.
    (a) Show that the equation whose roots are \frac {1}{\alpha} and \frac {1}{\beta} is x^2+6x+9=4kx
    (b) find the set of values of k for which \alpha and \beta are real and positive. Can anyone help me with this problem?
    HI

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

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

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

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

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

    =4k-6

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

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

    then expand it

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

    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 9x^2+6x+1 = 4kx where k is a real constant, are denoted by \alpha and \beta.
    (a) Show that the equation whose roots are \frac {1}{\alpha} and \frac {1}{\beta} is x^2+6x+9=4kx
    (b) find the set of values of k for which \alpha and \beta are real and positive. Can anyone help me with this problem?
    9x^2+6x+1 = 4kx
    9x^2+6x-4kx+1 =0
    9x^2+(6-4k)x+1 =0
    if 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.
    \alpha + \beta=\frac{-b}{a}=....
    and
    \alpha x \beta=\frac{c}{a}=....

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

    after you determine the sum and product, the equation would be:
    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

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

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

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

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

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

    =4k-6

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

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

    then expand it

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

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