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Thread: [SOLVED] How to solve this question?

  1. #1
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    [SOLVED] How to solve this question?

    the roots of the quadratic equation $\displaystyle x^2 + mx + n = 0$ are $\displaystyle \alpha$ and $\displaystyle \beta$. If the roots are in the ratio a:b, show that mab = n(a + b)
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  2. #2
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    Hello, Mark!

    The roots of the quadratic equation $\displaystyle x^2 + mx + n \:=\: 0$ are $\displaystyle \alpha$ and $\displaystyle \beta$.

    If the roots are in the ratio $\displaystyle a:b$, show that: .$\displaystyle m^2ab \:=\: n(a + b)^2$

    We have: .$\displaystyle \begin{Bmatrix}\alpha &=& ak \\ \beta &=& bk \end{Bmatrix}\;\; \text{ for some constant }k.$

    And we have: .$\displaystyle \begin{array}{cccccccc}
    \alpha + \beta &=& \text{-}m & \Longrightarrow & ak + bk &=& \text{-}m & {\color{blue}[1]} \\ \alpha\beta &=& n & \Longrightarrow & ak\cdot bk &=& n & {\color{blue}[2]} \end{array}$


    From [1], we have: .$\displaystyle (a+b)k \:=\:\text{-}m \quad\Rightarrow\quad k \:=\:\frac{\text{-}m}{a+b}\;\;{\color{blue}[3]}$

    From [2], we have: .$\displaystyle abk^2 \:=\:n\;\;{\color{blue}[4]} $


    Substitute [3] into [4]: .$\displaystyle ab\left(\frac{\text{-}m}{a+b}\right)^2 \:=\:n \quad\Rightarrow\quad ab\,\frac{m^2}{(a+b)^2} \:=\:n$


    . . Therefore: .$\displaystyle m^2ab \;=\;n(a+b)^2$

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