Pierwiastki x_1,x_2 równania 2(x+2)(x-2)=0 spełniają warunek

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

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

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

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