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Math Help - [SOLVED] How to solve this question?

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

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

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

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

    And we have: . \begin{array}{cccccccc}<br />
\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: . (a+b)k \:=\:\text{-}m \quad\Rightarrow\quad k \:=\:\frac{\text{-}m}{a+b}\;\;{\color{blue}[3]}

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


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


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

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