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Math Help - Roots of a quadratic equation .

  1. #1
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    Roots of a quadratic equation .

    Q : If a, b and c \in IR , with a is not 0 , and the roots of the equation ax^2+bx+c=0 are real , show that the roots of a^2y^2-(b^2-2ac)y+c^2=0 are also real .
    If the roots of the quadratic equation ax^2+bx+c=0 are \alpha and \beta , state the value of \alpha+\beta and \alpha\beta in terms of a ,b and c . Hence , find the roots of the second equation in terms of \alpha and \beta .

    I don understand this part only --- hence , find the roots of the second equation ( the last sentence )
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    Q : If a, b and c \in IR , with a is not 0 , and the roots of the equation ax^2+bx+c=0 are real , show that the roots of a^2y^2-(b^2-2ac)y+c^2=0 are also real .
    If the roots of the quadratic equation ax^2+bx+c=0 are \alpha and \beta , state the value of \alpha+\beta and \alpha\beta in terms of a ,b and c . Hence , find the roots of the second equation in terms of \alpha and \beta .

    I don understand this part only --- hence , find the roots of the second equation ( the last sentence )
    If the roots of ax^2+bx+c=0 are \alpha and \beta then

    ax^2+bx+c=0 \; \Rightarrow\; (x - \alpha)(x - \beta) = 0 which on expanding and comparing gives

     (1)\;\;\;\alpha + \beta = - \frac{b}{a}\; \text{and}\; \alpha\, \beta = \frac{c}{a}

    The roots of your second equation are given by

    (2)\;\;\;r_1, r_2 = \frac{b^2-2ac \pm \sqrt{(b^2-2ac)^2 - 4a^2c^2}}{2a^2} = \frac{b^2-2ac \pm \sqrt{b^4-4ab^2c}}{2a^2}

    From (1), isolate b\; \text{and}\;c and use these to eliminate them in (2). Simplify (the a's should cancel) and you get your result.
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  3. #3
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    Re :

    Thanks Danny , i understood fully what you did , but i encounter some problem in simplifying .

    \frac{b^2-2ac}{2a^2}+\frac{\sqrt{b^4-4ab^2c}}{2a^2}

    =\frac{1}{2}(\frac{b}{a})^2-\frac{c}{a}+\sqrt{\frac{1}{4}(\frac{b}{a})^4}-\frac{c}{a}(\frac{b}{a})^2

    Substituting one in here , i seemed to make the equation a little complicated .

    \frac{1}{2}(\alpha+\beta)^2-\alpha\beta+\sqrt{\frac{1}{4}(\alpha+\beta)^4-\alpha\beta(\alpha+\beta)^2}

    which brings me nowhere . Or there is a simpler way to do this . Thanks
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  4. #4
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    Hello, mathaddict!

    I think I understand what they want.
    I don't know if I've found the shortest way, though . . .


    If a, b,c \in IR,\;a \neq 0, and the roots of the equation ax^2+bx+c\:=\:0 are real,
    show that the roots of a^2y^2-(b^2-2ac)y+c^2\:=\:0 are also real.

    If the roots of the quadratic equation ax^2+bx+c\:=\:0 are \alpha, \beta,
    state the value of \alpha+\beta and \alpha\beta in terms of a ,b, c.

    Hence, find the roots of the second equation in terms of \alpha and \beta.
    We already know that: . \begin{Bmatrix}\alpha + \beta \:=\:-\dfrac{b}{a} \\ \alpha\beta \:=\:\dfrac{c}{a} \end{Bmatrix} . [1]


    Using the Quadratic Formula on the second equation,

    . . we have: . y \;=\;\frac{(b^2-2ac) \pm b\sqrt{b^2-4ac}}{2a^2}


    We will now manipulate this expression beyond all recognition . . .

    We have: . y \;=\;\frac{b^2}{2a^2} - \frac{2ac}{2a^2 } \pm\frac{b\sqrt{b^2-4ac}}{2a^2} . = \;\frac{1}{2}\!\cdot\!\frac{b^2}{a^2} - \frac{c}{a} \pm\frac{1}{2}\!\cdot\!\frac{b}{a}\!\cdot\!\frac{\  sqrt{b^2-4ac}}{a}

    . . . . . . . . y\;= \;\frac{1}{2}\left(\frac{b}{a}\right)^2 - \left(\frac{c}{a}\right) \pm \frac{1}{2}\left(\frac{b}{a}\right) \sqrt{\left(\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right)} . [2]


    Substitute [1] into [2]:

    . . y \;=\;\frac{1}{2}(\text{-}\alpha-\beta)^2 - (\alpha\beta) \:\pm \:\frac{1}{2}(\text{-}\alpha-\beta)\sqrt{(\text{-}\alpha-\beta)^2 - 4(\alpha\beta)}

    . . y \;=\;\frac{1}{2}(\alpha^2 + \beta^2) \:\mp\: \frac{1}{2}(\alpha + \beta)\sqrt{\alpha^2 - 2\alpha\beta + \beta^2}

    . . y \;=\;\frac{1}{2}(\alpha^2+\beta^2) \:\mp\: \frac{1}{2}(\alpha + \beta)(\alpha - \beta)

    . . y \;=\;\frac{1}{2}\bigg[(\alpha^2+\beta^2) \pm(\alpha^2-\beta^2)\bigg]


    Therefore: . y \;=\;\alpha^2,\;\beta^2


    The simplicity of these answers leads me to suspect that
    . . there is a more direct approach.
    So far, I haven't found it . . . anyone? anyone?


    Edit: Too fast for me, Danny!
    Last edited by Soroban; January 12th 2009 at 08:10 AM.
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  5. #5
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    Let \gamma and \delta be the roots of a^2y^2-(b^2-2ac)y+c^2=0

    Then

    \begin{Bmatrix}\gamma + \delta \:=\:\dfrac{b^2-2ac}{a^2} \\ \gamma\delta \:=\:\dfrac{c^2}{a^2} \end{Bmatrix}

    \dfrac{b^2-2ac}{a^2} = \left(\dfrac{b}{a}\right)^2 - 2 \:\dfrac{c}{a} = (\alpha + \beta)^2 - 2 \alpha \beta = \alpha^2 + \beta^2

    \dfrac{c^2}{a^2} = \left(\dfrac{c}{a}\right)^2 = (\alpha \beta)^2 = \alpha^2 \beta^2

    Then

    \gamma + \delta = \alpha^2 + \beta^2

    \gamma \delta = \alpha^2 \beta^2

    I am not sure this helps a lot
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  6. #6
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    Roots of a quadratic

    Quote Originally Posted by running-gag View Post
    \gamma + \delta = \alpha^2 + \beta^2

    \gamma \delta = \alpha^2 \beta^2

    I am not sure this helps a lot
    Isn't this exactly what Soroban was looking for? The roots are therefore \alpha^2 and \beta^2.

    Grandad
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  7. #7
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    Re :

    Yeah , running-gag found the shortest solution , haha , thanks all for helping me out .
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  8. #8
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    Quote Originally Posted by Grandad View Post
    Isn't this exactly what Soroban was looking for? The roots are therefore \alpha^2 and \beta^2.

    Grandad
    I was not completely sure that other pairs of \gamma and \delta could fit as well but no, you re right
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