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Thread: roots of an equation

  1. #1
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    roots of an equation

    The roots of an equation:

    px^2 + qx + 1 = 0 are r1 and r2

    Find r1^2 + r2^2 in terms of p and q
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  2. #2
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    Hello, DINOCALC09!

    The roots of the equation: .$\displaystyle px^2 + qx + 1 \:= \:0$ are $\displaystyle r_1\text{ and }r_2.$

    Find $\displaystyle r_1^2 + r_2^2$ in terms of $\displaystyle p\text{ and }q.$
    Divide by $\displaystyle p\!:\;\;x^2 + \frac{q}{p}x + \frac{1}{p} \;=\;0$


    There is a theorem that says:

    . . $\displaystyle [1]\;\;r_1+ r_2 \:=\:-\frac{q}{p}$ . . . The sum of the roots is the negative of the x-coefficient.

    . . $\displaystyle [2]\;\;r_1\!\cdot\!r_2 \:=\:\frac{1}{p}$ . . . The product of the roots is the constant term.


    Square [1]: .$\displaystyle (r_1+r_2)^2 \:=\:\left(-\frac{q}{p}\right)^2\quad\Rightarrow\quad r_1^2 + 2r_1r_2 + r_2^2 \:=\:\frac{q^2}{p^2}$

    Substitute [2]: .$\displaystyle r_1^2 + 2\left(\frac{1}{p}\right) + r_2^2 \;=\;\frac{q^2}{p^2} \quad\Rightarrow\quad r_1^2 + r_2^2 \;=\;\frac{q^2}{p^2} - \frac{2}{p}$

    . . Therefore: .$\displaystyle r_1^2 + r_2^2 \;=\;\frac{q^2-2p}{p^2}$

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