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Math Help - Construct equation.

  1. #1
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    Construct equation.

    If α, β are the roots of the equation x^2 - 5x + 3 = 0, construct an equation with roots α/ β^2 and β/ α^2.

    Help I'm having trouble with this q,i got -18/25 for the sum and 1/5 for product. But the answer is wrong. I think im calculating the product wrong when i multiply.
    Can someone show the steps to do this please. Thanks!
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  2. #2
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    Smile

    Quote Originally Posted by Detanon View Post
    If α, β are the roots of the equation x^2 - 5x + 3 = 0, construct an equation with roots α/ β^2 and β/ α^2.

    Help I'm having trouble with this q,i got -18/25 for the sum and 1/5 for product. But the answer is wrong. I think im calculating the product wrong when i multiply.
    Can someone show the steps to do this please. Thanks!
    let \mu = \frac{\alpha}{\beta^2} and \gamma = \frac{\beta}{\alpha^2} be the roots of the new polynomial, then

    \mu + \gamma = \frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2} =\frac{\alpha^3 + \beta^3}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)}{(\alpha \beta)^2}

    and

    \mu \gamma = \frac{\alpha}{\beta^2} \times \frac{\beta}{\alpha^2} = \frac{1}{\alpha \beta}

    where
    \alpha + \beta = 5 and \alpha \beta = 3

    the new polynomial is p(x) = x^2 - (\mu + \gamma)x + \mu \gamma
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  3. #3
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    Hello, Detanon!

    I'll get you started . . .


    \alpha, \beta are the roots of the equation x^2 - 5x + 3 \:=\: 0

    Construct an equation with roots \frac{\alpha}{\beta^2} and \frac{\beta}{\alpha^2}

    The new equation will have:

    . . x-coefficient: . b \;=\;-\left(\frac{\alpha}{\beta^2} + \frac{\beta}{\alpha^2}\right) \;=\;-\frac{\alpha^3 + \beta^3}{\alpha^2\beta^2}

    . . Constant term: . c \;=\;\left(\frac{\alpha}{\beta^2}\right)\left(\fra  c{\beta}{\alpha^2}\right) \;=\;\frac{\beta}{\alpha}



    We have: . \begin{array}{cccc}\alpha + \beta &=& 5 & [1] \\ \alpha\beta &=& 3 & [2] \end{array}


    Cube [1]: . (\alpha + \beta)^3 \:=\:5^3 \quad\Rightarrow\quad \alpha^3 + 3\alpha^2\beta + 3\alpha\beta^2 + \beta^3 \:=\:125

    . . \alpha^3 + \beta^3 + 3\underbrace{\alpha\beta}_{3}\underbrace{(\alpha + \beta)}_{5} \;=\;125 \quad\Rightarrow\quad \alpha^3+\beta^3 \:=\:80

    Square [2]: . (\alpha\beta)^2 \:=\:3^2 \quad\Rightarrow\quad \alpha^2\beta^2 \:=\:9


    Hence: . b \;=\;-\frac{\alpha^3 + \beta^3}{\alpha^2\beta^2} \:=\:-\frac{80}{9}

    . . and we've found the x-coefficient . . .

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