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Math Help - Roots of quadratic equation

  1. #1
    Member maybeline9216's Avatar
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    Roots of quadratic equation

    Please teach me how to do part b(iii).
    Attached Thumbnails Attached Thumbnails Roots of quadratic equation-benqn6.jpg  
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by maybeline9216 View Post
    Please teach me how to do part b(iii).
    note that (\alpha - \beta)^2 = \alpha^2 - 2 \alpha \beta  + \beta^2. with this, you can find \alpha - \beta by square rooting both sides.

    but how to find \alpha^2 - 2 \alpha \beta  + \beta^2?

    note that (\alpha + \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2

    \Rightarrow (\alpha + \beta)^2 - 4 \alpha \beta = \color{red}\alpha^2 - 2 \alpha \beta  + \beta^2

    and you know \alpha + \beta as well as \alpha \beta from part b(i)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by maybeline9216 View Post
    Please teach me how to do part b(iii).
    lets assume we are going to do this without using the quadratic formula.

    (\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha \beta=(\alpha+\beta)^2-4\alpha \beta

    so:

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

    and we know that \alpha+\beta=k=7 and that \alpha \beta=2k-3=11

    CB
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