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Math Help - Quadratic Equation Part 2

  1. #1
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    Quadratic Equation Part 2

    How to do these?

    1. The roots of the quadratic equation x^2+3x+1=0are \alphaand \beta. Find the equation when the roots are 2\alpha-\beta,2\beta-\alpha

    2. The roots of an equation are the reciprocals of the roots of the equation x^2+2ax-c^2=0. Find that equation.

    3. If one of the roots of the equation mx^2+nx+p=0is twice the other root, find the relation between m,n and p.

    Answer:
    1. x^2+3x-9=0

    2. c^2x^2-2ax-1=0

    3. 9pm= 2n^2
    Last edited by cloud5; June 24th 2009 at 06:41 AM.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    1. From the first quadratic equation, \alpha+\beta=-3
    \alpha\beta=1

    Let the new roots be \alpha' and \beta'

    \alpha'+\beta' = (2\alpha-\beta)+(2\beta-\alpha)

    = \alpha+\beta

    = -3

    \alpha'\beta' = (2\alpha-\beta)(2\beta-\alpha)

    = 4\alpha\beta-2{\beta}^2-2{\alpha}^2+\alpha\beta

    = 5\alpha\beta-2({\alpha}^2+{\beta}^2)

    = 5\alpha\beta-2({\alpha}^2+{\beta}^2+2{\alpha}{\beta})+4{\alpha}  {\beta}

    = 9\alpha\beta-2(\alpha+\beta)^2

    = 9-2(-3)^2

    = 9-18

    = -9

    The new quadratic equation can be written as x^2-x(\alpha'+\beta')+\alpha'\beta'=0
    x^2-x(-3)+(-9)=0

    x^2+3x-9=0
    -----------------------------------------------------------------------------------------------------

    2. Let the roots of the original equation be \alpha and \beta

    \alpha+\beta=-2a

    \alpha\beta=-c^2

    Let the new roots be \alpha' and \beta'

    \alpha'+\beta'=\frac{1}{\alpha}+\frac{1}{\beta}

    \alpha'+\beta'=\frac{{\alpha}+{\beta}}{\alpha\beta  }

    \alpha'+\beta'=\frac{-2a}{-c^2}

    \alpha'+\beta'=\frac{2a}{c^2}

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

    \alpha'\beta'=\frac{-1}{c^2}

    Let the new roots be \alpha' and \beta'

    The new quadratic equation can be written as x^2-x(\alpha'+\beta')+\alpha'\beta'=0

    x^2-x(\frac{2a}{c^2})+\frac{-1}{c^2}=0

    c^2x^2-2ax-1=0
    ---------------------------------------------------------------------------------------------------------

    3. Let the roots be \alpha and 2\alpha

    \alpha+2\alpha=-\frac{n}{m}

    3\alpha=-\frac{n}{m}

    9{\alpha}^2=\frac{n^2}{m^2} -------------(1)

    \alpha*2\alpha=\frac{p}{m}

    2{\alpha}^2=\frac{p}{m} ------------ (2)

    Equating {\alpha}^2 from equations (1) and (2),

    \frac{n^2}{9m^2}=\frac{p}{2m}

    \frac{n^2}{9m}=\frac{p}{2}

    9pm=2n^2
    Last edited by alexmahone; June 24th 2009 at 05:31 PM.
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  3. #3
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    Hello, cloud5!

    We're expected to know this . . .

    Given the quadratic equation: . x^2 + px + q \:=\:0 . .
    (The leading coefficient is 1.)

    . . the roots \alpha\text{ and }\beta are such that: . \begin{Bmatrix}\alpha + \beta &=& -p \\ \alpha\beta &=& q \end{Bmatrix}

    The sum of the roots is the negative of the x-coefficient.
    The product of the roots is the constant term.



    1. The roots of the quadratic equation x^2+3x+1\:=\:0 are \alpha and \beta.

    Find the equation when the roots are: 2\alpha-\beta,\;2\beta-\alpha

    Answer: . x^2+3x-9\:=\:0
    There is a dazzlingly clever method around all the expected algebra . . .


    The new equation has roots: . 2\alpha-\beta\text{ and }2\beta-\alpha

    . . Its x-coefficient will be: . p \;=\;-\bigg[(2\alpha-\beta) + (2\beta - \alpha)\bigg] \;=\;-(\alpha + \beta) .(a)

    . . The constant term will be: . q \;=\;-(2\alpha-\beta)(2\beta-\alpha) \:=\:-2(\alpha^2 + \beta^2) + 5(\alpha\beta) .(b)


    From the original equation, we have: . \begin{Bmatrix}\alpha + \beta &=& \text{-}3 & {\color{blue}[1]}\; \\ \alpha\beta &=& 1 & {\color{blue}[2]\;} \end{Bmatrix}

    Substitute [1] into (a): . p \;=\;-(-3) \quad\Rightarrow\quad \boxed{p \:=\:3}


    \text{Square }{\color{blue}[1]}\!:\quad (\alpha + \beta)^2 \:=\:(-3)^2 \quad\Rightarrow\quad (\alpha^2 + \beta^2) + 2\!\!\underbrace{(\alpha\beta)}_{\text{This is 1}} \:=\:9

    . . Hence: . (\alpha^2+\beta^2) + 2 \:=\:9 \quad\Rightarrow\quad \alpha^2 + \beta^2 \:=\:7\;\;{\color{blue}[3]}

    Substitute [2] and [3] into (b): . q \;=\;-2(7) + 5(1) \quad\Rightarrow\quad \boxed{q\;=\;-9}


    Therefore, the new equation is: . \boxed{x^2 + 3x - 9 \:=\:0}




    2. The roots of an equation are the reciprocals of the roots of: . x^2+2ax-c^2\:=\:0

    Find that equation.

    Answer: . c^2x^2 - 2ax - 1 \:=\:0
    The roots of the original equation are: . \alpha\text{ and }\beta

    . . So we have: . \begin{Bmatrix}\alpha + \beta &=& \text{-}2a & {\color{blue}[1]} \\ \alpha\beta &=& \text{-}c^2 & {\color{blue}[2]} \end{Bmatrix}


    The roots of the new equation are: . \frac{1}{\alpha}\text{ and }\frac{1}{\beta}

    Its x-coefficient will be: . p \;=\;-\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \;=\;-\left(\frac{\alpha+\beta}{\alpha\beta}\right)

    . . Substitute [1] and [2]: . p \:=\:-\left(\frac{\text{-}2a}{\text{-}c^2}\right) \quad\Rightarrow\quad\boxed{ p \;=\;-\frac{2a}{c^2}}

    Its constant term will be: . q \;=\;\frac{1}{\alpha}\!\cdot\!\frac{1}{\beta} \;=\;\frac{1}{\alpha\beta}

    . . Substitute [2]: . q \;=\;\frac{1}{\text{-}c^2} \quad\Rightarrow\quad\boxed{ q \;=\;-\frac{1}{c^2}}


    The new equation is: . x^2 - \frac{2a}{c^2}x - \frac{1}{c^2} \;=\;0

    Multiply by c^2\!:\quad \boxed{c^2x^2 - 2ax - 1 \;=\;0}



    Care to try the third one youself?

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