Question
Solve: $\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
✓
Answer
$\sqrt3\text{x}^2+10\text{x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}^2+\text{12x}-\text{2x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+4\sqrt3\big)\big(\sqrt3\text{x}-2\big)=0$
$\Rightarrow\text{x}+4\sqrt3=0$ or $\sqrt3\text{x}-2=0$
$\Rightarrow\text{x}=-4\sqrt3$ or $\text{x}=\frac{2}{\sqrt3}$
$\Rightarrow\sqrt3\text{x}^2+\text{12x}-\text{2x}-8\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+4\sqrt3\big)-2\big(\text{x}+4\sqrt3\big)=0$
$\Rightarrow\big(\text{x}+4\sqrt3\big)\big(\sqrt3\text{x}-2\big)=0$
$\Rightarrow\text{x}+4\sqrt3=0$ or $\sqrt3\text{x}-2=0$
$\Rightarrow\text{x}=-4\sqrt3$ or $\text{x}=\frac{2}{\sqrt3}$
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