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Math Help - Difficult quadratic equation problems

  1. #1
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    Difficult quadratic equation problems

    Hi,

    How do i solve these?

    1)
    If \alpha and \beta are the roots of the equation ax^2 + bx + c = 0, then the equation whose roots are \alpha + (1/\beta) and \beta + (1/\alpha), is?

    2)
    If the equations 2x^2 + 3x + 5\gamma = 0 and x^2 + 2x + 3\gamma = 0 have a common root, then \gamma = ?

    3)
    If the difference of the roots of the equations x^2 + bx + c = 0 be 1, then:

    (a) b^2-4c-1 = 0
    (b) b^2 - 4c = 0
    (c) b^2 - 4c + 1 = 0
    (d) b ^2 + 4c - 1 = 0
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  2. #2
    Moo
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    Hello,

    Let's use brute force... for the first question :

    If \alpha,\beta are the roots, then \alpha+\beta=-\frac ba and \alpha\beta=\frac ca

    Let x^2+\mu x+\nu=0 be the equation which roots are \alpha+\frac 1\beta,\beta+\frac 1\alpha

    Then -\mu=\alpha+\tfrac 1\beta+\beta+\tfrac 1\alpha=(\alpha+\beta)+\tfrac{\alpha+\beta}{\alpha  \beta}=-\tfrac ba-\tfrac bc
    So \mu=\tfrac ba+\tfrac bc

    And \nu=\left(\alpha+\tfrac 1\beta\right)\cdot\left(\beta+\tfrac 1\alpha\right)=\alpha\beta+\tfrac{1}{\alpha\beta}+  2=\tfrac ca+\tfrac ac+2

    So the equation you're looking for is x^2+\left(\tfrac ba+\tfrac bc\right)x+\tfrac ca+\tfrac ac+2=0 \Leftrightarrow acx^2+b(a+c)x+(a+c)^2=0
    Last edited by Moo; December 29th 2009 at 09:40 AM.
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  3. #3
    Moo
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    For question 2)...

    There exists \delta such that :

    \begin{cases} 2\delta^2+3\delta+5\gamma=0 \\ \delta^2+2\delta+3\gamma=0\end{cases}

    You may be able to solve for \gamma by yourself eh ?


    For question 3)...

    We know that the sum of the two roots \alpha,\beta is -b and their product is c.
    So we have b^2=(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta  =\alpha^2+\beta^2+2c

    Hence \alpha^2+\beta^2=b^2-2c

    But 1=(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta=\alpha^2+\beta^2-2c=b^2-4c

    Finally, b^2-4c-1=0
    Last edited by Moo; December 29th 2009 at 09:40 AM.
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  4. #4
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    Hello, saberteeth!

    My approach to #1 is similar to Moo's . . . with a different answer.
    . . Did I mess up?


    1) If \alpha and \beta are the roots of the equation: ax^2 + bx + c \:=\: 0,
    find the equation whose roots are: . \alpha + \tfrac{1}{\beta} and \beta + \tfrac{1}{\alpha} . . . . in terms of {\color{blue}a,b,c}

    Since \alpha,\:\beta are roots of: . x^2 + \frac{b}{c}x + \frac{c}{a} \:=\:0
    . . then: . \begin{array}{cccc}\alpha + \beta &=& \text{-}\dfrac{b}{a} & [1] \\ \\[-3mm] \alpha\beta &=& \dfrac{c}{a} & [2] \end{array}


    The sum of the two roots is:

    . . S \;=\;\left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) \;=\; (\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) . =\;(\alpha + \beta) + \left(\frac{\alpha + \beta}{\alpha\beta}\right) \;=\;(\alpha+\beta)\left(1 + \frac{1}{\alpha\beta}\right)

    Substitute [1] and [2]: . S \;=\;-\frac{b}{a}\left(1 + \frac{1}{\frac{c}{a}}\right) \;=\; -\frac{b(a+c)}{ac}

    . . Hence, the x-coefficient is: . \frac{b(a+c)}{ac}



    The product of the two roots is:

    . . P \;=\;\left(\alpha + \frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}  \right) \;=\;\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}

    Substitute [2]: . P \;=\;\frac{c}{a} + 2 + \frac{a}{c} \;=\;\frac{a^2 + 2ac + c^2}{ac}

    . . Hence, the constant term is: . \frac{(a+c)^2}{ac}


    Therefore, the equation is: . x^2 + \frac{b(a+c)}{ac}x + \frac{(a+c)^2}{ac} \:=\:0 \quad\Rightarrow\quad\boxed{ acx^2 + b(a+c)x + (a+c)^2 \:=\:0}

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  5. #5
    Moo
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    Yep Soroban, a slight error in the signs and an incorrect factorisation
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