Question
Solve the following quadratic equation:$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$
✓
Answer
$4\sqrt6\text{x}^2-\text{13x}-2\sqrt6=0$$\Rightarrow4\sqrt6\text{x}^2-16\text{x}+3\text{x}-2\sqrt6=0$
$\Rightarrow4\sqrt2\text{x}\big(\sqrt3\text{x}-2\sqrt2\big)+\sqrt3\big(\sqrt3\text{x}-2\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-2\sqrt2\big)\big(4\sqrt2\text{x}+\sqrt3\big)=0$
$\Rightarrow\Big(\frac{2\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\Big)=\frac{2\sqrt6}{3}$ or $\Big(\frac{-\sqrt3}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2}\Big)=\frac{-\sqrt6}{8}$
Hence, $\frac{2\sqrt6}{3}$ and $\frac{-\sqrt6}{8}$ are the roots of the given equation.
$\Rightarrow4\sqrt2\text{x}\big(\sqrt3\text{x}-2\sqrt2\big)+\sqrt3\big(\sqrt3\text{x}-2\sqrt2\big)=0$
$\Rightarrow\big(\sqrt3\text{x}-2\sqrt2\big)\big(4\sqrt2\text{x}+\sqrt3\big)=0$
$\Rightarrow\Big(\frac{2\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\Big)=\frac{2\sqrt6}{3}$ or $\Big(\frac{-\sqrt3}{4\sqrt2}\times\frac{\sqrt2}{\sqrt2}\Big)=\frac{-\sqrt6}{8}$
Hence, $\frac{2\sqrt6}{3}$ and $\frac{-\sqrt6}{8}$ are the roots of the given equation.
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