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Thread: Difficult quadratic equation problems

  1. #1
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    Difficult quadratic equation problems

    Hi,

    How do i solve these?

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

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

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

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

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

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

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

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

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

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

    There exists $\displaystyle \delta$ such that :

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

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


    For question 3)...

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

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

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

    Finally, $\displaystyle b^2-4c-1=0$
    Last edited by Moo; Dec 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 $\displaystyle \alpha$ and $\displaystyle \beta$ are the roots of the equation: $\displaystyle ax^2 + bx + c \:=\: 0$,
    find the equation whose roots are: .$\displaystyle \alpha + \tfrac{1}{\beta}$ and $\displaystyle \beta + \tfrac{1}{\alpha}$ . . . . in terms of $\displaystyle {\color{blue}a,b,c}$

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


    The sum of the two roots is:

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

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

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



    The product of the two roots is:

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

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

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


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

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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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