g)
\( \displaystyle 2x^{2}-\frac{3}{2}x+\frac{1}{4}=0 \)
Calculando Δ
\[ \bbox[#99CCFF,10px]
{\Delta =b^{2}-4ac}
\]
\[
\begin{align}
& \Delta=\left(-{\frac{3}{2}}\right)^{2}-4.2.\frac{1}{4}\\
& \Delta=\frac{9}{4}-2=\frac{9-8}{4}\\
& \Delta =\frac{1}{4}
\end{align}
\]
Cálculo das raízes
\[ \bbox[#99CCFF,10px]
{x_{1,2}=\frac{-b\pm \sqrt{\Delta \;}}{2a}}
\]
\[
\begin{align}
& x_{1,2}=\frac{-\left(-{\dfrac{3}{2}}\right)\pm\sqrt{\dfrac{1}{4}\;}}{2.2}\\[5pt]
& x_{1,2}=\left(\frac{3}{2}\pm\frac{1}{2}\right).\frac{1}{4}\\[5pt]
& x_{1}=\left(\frac{3}{2}+\frac{1}{2}\right).\frac{1}{4}=\frac{\cancel{4}}{2}.\frac{1}{\cancel{4}}=\frac{1}{2}\\[5pt]
& x_{2}=\left(\frac{3}{2}-\frac{1}{2}\right).\frac{1}{4}=\frac{\cancel{2}}{\cancel{2}}.\frac{1}{4}=\frac{1}{4}
\end{align}
\]
\[ \bbox[#FFCCCC,10px]
{V=\left\{\frac{1}{2},\frac{1}{4}\right\}}
\]