Question
Solve the following equation by factorization$4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0$
✓
Answer
$
\begin{aligned}
& 4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0 \\
& \{4 \sqrt{3} \times(-2 \sqrt{3})=8 \times(-3)=-24\} \\
& 4 \sqrt{3} x^2+8 x-3 x-2 \sqrt{3}=0 \\
& \Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0 \\
& \Rightarrow(\sqrt{3} x+2)(4 x-\sqrt{3})=0
\end{aligned}
$
Either $\sqrt{3} x+2=0$,
then $\sqrt{3} x=-2$
$
\begin{aligned}
& \Rightarrow x =-\frac{2}{\sqrt{3}} \\
& \Rightarrow x =\frac{-2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-2 \sqrt{3}}{3}
\end{aligned}
$
or
$
4 x-\sqrt{3}=0 \text {, }
$
then $4 x=\sqrt{3}$
$
\Rightarrow x =\frac{\sqrt{3}}{4}
$
Hence $x=\frac{-2 \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$.
\begin{aligned}
& 4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0 \\
& \{4 \sqrt{3} \times(-2 \sqrt{3})=8 \times(-3)=-24\} \\
& 4 \sqrt{3} x^2+8 x-3 x-2 \sqrt{3}=0 \\
& \Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0 \\
& \Rightarrow(\sqrt{3} x+2)(4 x-\sqrt{3})=0
\end{aligned}
$
Either $\sqrt{3} x+2=0$,
then $\sqrt{3} x=-2$
$
\begin{aligned}
& \Rightarrow x =-\frac{2}{\sqrt{3}} \\
& \Rightarrow x =\frac{-2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-2 \sqrt{3}}{3}
\end{aligned}
$
or
$
4 x-\sqrt{3}=0 \text {, }
$
then $4 x=\sqrt{3}$
$
\Rightarrow x =\frac{\sqrt{3}}{4}
$
Hence $x=\frac{-2 \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$.
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