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Thread: Roots

  1. #1
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    Roots

    Hi,

    If $\displaystyle x_{1}, x_{2} $ are the roots of $\displaystyle ax^2 +bx+c = 0$, find the value of
    $\displaystyle (ax_{1}+b)^{-2} +(ax_{2}+b)^{-2} $

    Thank you in advanced
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  2. #2
    Moo
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    Hello !

    I think I figured it out...
    Maybe there is a quicker way, but I have not a lof of time.

    Quote Originally Posted by Recklessid View Post
    Hi,

    If $\displaystyle x_{1}, x_{2} $ are the roots of $\displaystyle ax^2 +bx+c = 0$, find the value of
    $\displaystyle N=(ax_{1}+b)^{-2} +(ax_{2}+b)^{-2} $

    Thank you in advanced
    Remember the sum of the roots & the product of the roots :

    $\displaystyle x_1+x_2=-\frac ba \quad (1) \quad \quad \quad x_1x_2=\frac ca \quad (2)$


    $\displaystyle N=\frac{1}{(ax_1+b)^2}+\frac{1}{(ax_2+b)^2}$

    From (1), we know that $\displaystyle x_1=-\frac ba-x_2 \quad \Rightarrow \quad ax_1=-b-ax_2$.
    Similarly, $\displaystyle ax_2=-b-ax_1$.

    Substituting in N :

    $\displaystyle \begin{aligned} N&=\frac{1}{(-b-ax_2+b)^2}+\frac{1}{(-b-ax_1+b)^2} \\ \\
    &=\frac{1}{(ax_2)^2}+\frac{1}{(ax_1)^2} \\ \\
    &=\frac{1}{a^2} \cdot \left(\frac{1}{x_2^2}+\frac{1}{x_1^2}\right) \end{aligned}$


    Gathering it in a unique fraction :

    $\displaystyle N=\frac{1}{a^2} \cdot \frac{x_1^2+x_2^2}{x_1^2 \cdot x_2^2}$

    Completing the square above :

    $\displaystyle N=\frac{1}{a^2} \cdot \frac{({\color{red}x_1+x_2})^2-2{\color{blue}x_1x_2}}{({\color{blue}x_1x_2})^2}$

    But $\displaystyle {\color{blue}x_1x_2}=\frac ca$ and $\displaystyle {\color{red}x_1+x_2}=-\frac ba$.


    This simplifies into :

    $\displaystyle N=\frac{1}{a^2} \cdot \frac{\left(\frac ba\right)^2-2 \frac ca}{\left(\frac ca\right)^2}$

    $\displaystyle N=\frac{1}{\bold{a^2}} \cdot \left(\frac{b^2}{a^2}-2 \frac ca\right) \cdot \frac{\bold{a^2}}{c^2}$

    $\displaystyle N=\frac{1}{c^2} \cdot \left(\frac{b^2}{a^2}-2 \frac{ac}{a^2}\right)$

    $\displaystyle \boxed{N=\frac{b^2-2ac}{(ac)^2}}$


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  3. #3
    MHF Contributor red_dog's Avatar
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    Another solution:
    $\displaystyle x_1\neq 0, \ x_2\neq 0$ because, if $\displaystyle x_1=0$ then $\displaystyle c=0$ and $\displaystyle ax_2+b=0$, so the denominator is 0, contradiction.
    Now, if $\displaystyle x_1, \ x_2$ are the roots of the equation, then
    $\displaystyle ax_1^2+bx_1+c=0$
    $\displaystyle ax_2^2+bx_2+c=0$
    Divide the first equality by $\displaystyle x_1$ and the second by $\displaystyle x_2$:
    $\displaystyle ax_1+b=-\frac{c}{x_1}$
    $\displaystyle ax_2+b=-\frac{c}{x_2}$
    Then the expression becomes
    $\displaystyle \displaystyle\left(\frac{x_1}{c}\right)^2+\left(\f rac{x_2}{c}\right)^2=\frac{x_1^2+x_2^2}{c^2}=\frac {(x_1+x_2)^2-2x_1x_2}{c^2}=\frac{\frac{b^2}{a^2}-\frac{2c}{a}}{a^2}=\frac{b^2-2ac}{a^2c^2}$
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