Weze   x1  en  x2   de twee  oplossingen  van de vierkantsvergelijking
ax² + bx + c = 0 .
Dan is   \(\frac{1}{x_1}+\frac{1}{x_2} =\) 		A.   \(-\frac ba\)
B.   \(-\frac ab\)
C.    \(\frac ac\)
D.    \(\frac ab\)
E.    \(-\frac bc\)
    a    b    c    d    e

[ 4-0997 - op net sinds 11.10.2024-(E)-12.10.2024 ]

Translation in   E N G L I S H

IN CONSTRUCTION

Oplossing - Solution

\(\large\frac{1}{x_1}+\frac{1}{x_2} = \frac{x_1+x_2}{x_1.x_2}=\frac{-\frac ba}{\frac ca}=-\frac ba.\frac ac = -\frac bc\)
GWB