Question
Solve the following quadratic equation:$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$
✓
Answer
$\sqrt7\text{x}^2-\text{6x}-13\sqrt7=0$$\Rightarrow\sqrt7\text{x}^2-13\text{x}+7\text{x}-13\sqrt7=0$
$\Rightarrow\text{x}\big(\sqrt7\text{x}-13\big)+\sqrt7\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\big(\text{x}+\sqrt7\big)\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\text{x}+\sqrt7=0$ or $\sqrt7\text{x}-13=0$
$\Rightarrow\text{x}=-\sqrt7$ or $\text{x}=\frac{13\sqrt7}{7}$
Hence, $-\sqrt7$ and $\frac{13\sqrt7}{7}$ are the roots of the given equation.
$\Rightarrow\text{x}\big(\sqrt7\text{x}-13\big)+\sqrt7\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\big(\text{x}+\sqrt7\big)\big(\sqrt7\text{x}-13\big)=0$
$\Rightarrow\text{x}+\sqrt7=0$ or $\sqrt7\text{x}-13=0$
$\Rightarrow\text{x}=-\sqrt7$ or $\text{x}=\frac{13\sqrt7}{7}$
Hence, $-\sqrt7$ and $\frac{13\sqrt7}{7}$ are the roots of the given equation.
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