\require{AMSmath} Uitwerkingen Opgave 1 6x2+5x-6=0a=6, b=5 en c=-6D=52-4·6·-6=25+144=169$\begin{array}{l} x_{1,2} = \frac{{ - 5 \pm \sqrt {169} }}{{2 \cdot 6}} = \frac{{ - 5 \pm 13}}{{12}} \\ x = \frac{{ - 5 - 13}}{{12}} \vee \,x = \frac{{ - 5 + 13}}{{12}} \\ x = - \frac{{18}}{{12}} \vee x = \frac{8}{{12}} \\ x = - 1\frac{1}{2} \vee x = \frac{2}{3}\, \\ \end{array}$ 2x2-3x=22x²-3x-2=0a=2, b=-3 en c=-2D=(-3)²-4·2·-2=9+16=25$\begin{array}{l} x_{1,2} = \frac{{ - - 3 \pm \sqrt {25} }}{{2 \cdot 2}} = \frac{{3 \pm 5}}{4} \\ x = \frac{{ - 2}}{4} \vee x = \frac{8}{4} \\ x = - \frac{1}{2} \vee x = 2 \\ \end{array}$ 2x2=9x+52x²-9x-5=0a=2, b=-9 en c=-5D=(-9)²-4·2·-5=81+40=121$\begin{array}{l} x_{1,2} = \frac{{ - - 9 \pm \sqrt {121} }}{{2 \cdot 2}} = \frac{{9 \pm 11}}{4} \\ x = \frac{{ - 2}}{4} \vee x = \frac{{20}}{4} \\ x = - \frac{1}{2} \vee x = 5 \\ \end{array}$ ©2004-2024 WisFaq
\require{AMSmath}
Opgave 1 6x2+5x-6=0a=6, b=5 en c=-6D=52-4·6·-6=25+144=169$\begin{array}{l} x_{1,2} = \frac{{ - 5 \pm \sqrt {169} }}{{2 \cdot 6}} = \frac{{ - 5 \pm 13}}{{12}} \\ x = \frac{{ - 5 - 13}}{{12}} \vee \,x = \frac{{ - 5 + 13}}{{12}} \\ x = - \frac{{18}}{{12}} \vee x = \frac{8}{{12}} \\ x = - 1\frac{1}{2} \vee x = \frac{2}{3}\, \\ \end{array}$ 2x2-3x=22x²-3x-2=0a=2, b=-3 en c=-2D=(-3)²-4·2·-2=9+16=25$\begin{array}{l} x_{1,2} = \frac{{ - - 3 \pm \sqrt {25} }}{{2 \cdot 2}} = \frac{{3 \pm 5}}{4} \\ x = \frac{{ - 2}}{4} \vee x = \frac{8}{4} \\ x = - \frac{1}{2} \vee x = 2 \\ \end{array}$ 2x2=9x+52x²-9x-5=0a=2, b=-9 en c=-5D=(-9)²-4·2·-5=81+40=121$\begin{array}{l} x_{1,2} = \frac{{ - - 9 \pm \sqrt {121} }}{{2 \cdot 2}} = \frac{{9 \pm 11}}{4} \\ x = \frac{{ - 2}}{4} \vee x = \frac{{20}}{4} \\ x = - \frac{1}{2} \vee x = 5 \\ \end{array}$
Opgave 1
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