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Thread: 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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  2. #2
    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 $\displaystyle (\alpha - \beta)^2 = \alpha^2 - 2 \alpha \beta + \beta^2$. with this, you can find $\displaystyle \alpha - \beta$ by square rooting both sides.

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

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

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

    and you know $\displaystyle \alpha + \beta$ as well as $\displaystyle \alpha \beta$ 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.

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

    so:

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

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

    CB
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