pierwiastki-x1x2-rownania-2x2x-20-spelniaja-warunek

A)

$\displaystyle \frac{1}{x_1}+ \frac{1}{x_2}=-1$ \displaystyle \frac{1}{x_1}+ \frac{1}{x_2}=-1

B)

$\displaystyle \frac{1}{x_1}+ \frac{1}{x_2}=0$ \displaystyle \frac{1}{x_1}+ \frac{1}{x_2}=0

C)

$\displaystyle \frac{1}{x_1}+ \frac{1}{x_2}= \frac{1}{4}$ \displaystyle \frac{1}{x_1}+ \frac{1}{x_2}= \frac{1}{4}

D)

$\displaystyle \frac{1}{x_1}+ \frac{1}{x_2}= \frac{1}{2}$ \displaystyle \frac{1}{x_1}+ \frac{1}{x_2}= \frac{1}{2}