Question
Solve the following quadratic equation:$\text{x}^2-3\sqrt5\text{x}+10=0$
✓
Answer
$\text{x}^2-3\sqrt5\text{x}+10=0$$\Rightarrow\text{x}^2-\sqrt5\text{x}-2\sqrt{5}\text{x}+10=0$
$\Rightarrow\text{x}\big(\text{x}-\sqrt5\big)-2\sqrt5\big(\text{x}-\sqrt5\big)=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}-2\sqrt5\big)=0$
$\Rightarrow\text{x}-\sqrt5=0$ or $\text{x}-2\sqrt5=0$
$\Rightarrow\text{x}=\sqrt{5}$ or $\text{x}=2\sqrt5$
$\Rightarrow\text{x}\big(\text{x}-\sqrt5\big)-2\sqrt5\big(\text{x}-\sqrt5\big)=0$
$\Rightarrow\big(\text{x}-\sqrt5\big)\big(\text{x}-2\sqrt5\big)=0$
$\Rightarrow\text{x}-\sqrt5=0$ or $\text{x}-2\sqrt5=0$
$\Rightarrow\text{x}=\sqrt{5}$ or $\text{x}=2\sqrt5$
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