Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen
\(\frac{1}{12}x^2+\frac{1}{4}x-\frac{3}{2}=0\)
\(\frac{1}{12}x^2+\frac{1}{4}x-\frac{3}{2}=0\\
\Leftrightarrow \color{red}{12.} \left(\frac{1}{12}x^2+\frac{1}{4}x-\frac{3}{2}\right)=0 \color{red}{.12} \\
\Leftrightarrow x^2+3x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+3x-18=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (3)^2-4.1.(-18) & &\\
& = 9+72 & & \\
& = 81 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-3-\sqrt81}{2.1} & & = \frac{-3+\sqrt81}{2.1} \\
& = \frac{-12}{2} & & = \frac{6}{2} \\
& = -6 & & = 3 \\ \\ V &= \Big\{ -6 ; 3 \Big\} & &\end{align} \\ -----------------\)