Question
Solve the following quadratic equations by factorization:
$4\sqrt{3}\text{x}^2+5\text{x}-2\sqrt{3}=0$
$4\sqrt{3}\text{x}^2+5\text{x}-2\sqrt{3}=0$
✓
Answer
We have been given
$4\sqrt{3}\text{x}^2+5\text{x}-2\sqrt{3}=0$
$4\sqrt{3}\text{x}^2+8\text{x}-3\text{x}-2\sqrt{3}=0$
$4\text{x}\big(\sqrt{3}\text{x}+2\big)-\sqrt{3}\big(\sqrt{3}\text{x}+2\big)=0$
$\big(\sqrt{3}\text{x}+2\big)\big(4\text{x}-\sqrt{3}\big)=0$
Therefore,
$\sqrt{3}\text{x}+2=0$
$\sqrt{3}\text{x}=-2$
$\text{x}=\frac{-2}{\sqrt{3}}$
or, $4\text{x}-\sqrt{3}=0$
$4\text{x}=\sqrt{3}$
$\text{x}=\frac{\sqrt{3}}{4}$
Hence, $\text{x}=\frac{-2}{\sqrt{3}}$ or $\text{x}=\frac{\sqrt{3}}{4}$
$4\sqrt{3}\text{x}^2+5\text{x}-2\sqrt{3}=0$
$4\sqrt{3}\text{x}^2+8\text{x}-3\text{x}-2\sqrt{3}=0$
$4\text{x}\big(\sqrt{3}\text{x}+2\big)-\sqrt{3}\big(\sqrt{3}\text{x}+2\big)=0$
$\big(\sqrt{3}\text{x}+2\big)\big(4\text{x}-\sqrt{3}\big)=0$
Therefore,
$\sqrt{3}\text{x}+2=0$
$\sqrt{3}\text{x}=-2$
$\text{x}=\frac{-2}{\sqrt{3}}$
or, $4\text{x}-\sqrt{3}=0$
$4\text{x}=\sqrt{3}$
$\text{x}=\frac{\sqrt{3}}{4}$
Hence, $\text{x}=\frac{-2}{\sqrt{3}}$ or $\text{x}=\frac{\sqrt{3}}{4}$
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