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Math Help - Roots of an equation

  1. #1
    MHF Contributor alexmahone's Avatar
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    Roots of an equation

    If \alpha, \beta are the roots of
    (x - 2)(x - \sqrt3) = \sqrt7,
    find the roots of
    (x - \alpha)(x - \beta) + \sqrt7 = 0
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  2. #2
    MHF Contributor red_dog's Avatar
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    If \alpha, \ \beta are the roots of equation (x-2)(x-\sqrt{3})-\sqrt{7}=0 then

    (x-2)(x-\sqrt{3})-\sqrt{7}=(x-\alpha)(x-\beta)

    Then, (x-\alpha)(x-\beta)+\sqrt{7}=(x-2)(x-\sqrt{3})-\sqrt{7}+\sqrt{7}=

    =(x-2)(x-\sqrt{3})=0\Rightarrow x_1=2, \ x_2=\sqrt{3}
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  3. #3
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    Hello, alexmahone!


    I love your solution, red_dog!

    Here's my approach . . .


    If \alpha, \beta are the roots of: . (x - 2)(x - \sqrt3) \:=\: \sqrt7,

    . . find the roots of: . (x - \alpha)(x - \beta) + \sqrt7 \:=\: 0

    If \alpha\text{ and }\beta roots of: . x^2 - (2+\sqrt{3})x + (2\sqrt{3}-\sqrt{7}) \:=\:0

    . . then: . \begin{array}{cccc}\alpha + \beta &=& 2+\sqrt{3} & {\color{blue}[1]} \\ \\[-4mm]\alpha\beta &=& 2\sqrt{3}-\sqrt{7} & {\color{blue}[2]}\end{array}


    We want the roots of: . x^2 - (\alpha + \beta)x + (\alpha\beta + \sqrt{7}) \:=\:0

    Substitute [1] and [2]: . x^2 -(2+\sqrt{3})x + [(2\sqrt{3}-\sqrt{7}) + \sqrt{7}] \:=\:0

    . . and we have: . x^2 - (2 + \sqrt{3})x + 2\sqrt{3} \:=\:0


    The roots of this equation are p\text{ and }q where: . \begin{Bmatrix}p+q &=& 2+\sqrt{3} \\ pq &=& 2\sqrt{3} \end{Bmatrix}

    Therefore: . p \,=\,2,\;q \,=\,\sqrt{3}

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