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
Hello, DINOCALC09!
Divide by $\displaystyle p\!:\;\;x^2 + \frac{q}{p}x + \frac{1}{p} \;=\;0$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 $\displaystyle p\text{ and }q.$
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: .$\displaystyle r_1^2 + r_2^2 \;=\;\frac{q^2-2p}{p^2}$