Question
Solve the following quadratic equations by factorization:
$3\sqrt{5}\text{x}^2+25\text{x}-10\sqrt{5}=0$
$3\sqrt{5}\text{x}^2+25\text{x}-10\sqrt{5}=0$
✓
Answer
$3\sqrt{5}\text{x}^2+25\text{x}-10\sqrt{5}=0$
$\sqrt{5}\bigg(3\text{x}^2+\frac{25}{\sqrt{5}}\text{x}-\frac{10\sqrt{5}}{\sqrt{5}}\bigg)=0$
$\Rightarrow\sqrt{5}\big[3\text{x}^2+5\sqrt{5}\text{x}-10\big]=0$
$\big(\text{Dividing by}\sqrt{5}\big)$
$\Rightarrow\big[3\text{x}^2-\sqrt{5}\text{x}+6\sqrt{5}\text{x}-10\big]=0$
$\Rightarrow3\text{x}^2-\sqrt{5}\text{x}+6\sqrt{5}\text{x}-10=0$
$\Rightarrow\text{x}\big(3\text{x}-\sqrt{5}\big)+2\sqrt{5}\big(3\text{x}-\sqrt{5}\big)=0$
$\Rightarrow\big(3\text{x}-\sqrt{5}\big)\big(\text{x}+2\sqrt{5}\big)=0$
Either $3\text{x}-\sqrt{5}=0,$ then $3\text{x}=\sqrt{5}$
$\Rightarrow\text{x}=\frac{\sqrt{5}}{3}$
or $\text{x}+2\sqrt{5}=0,$ then $\text{x}=-2\sqrt{5}$
$\therefore\text{x}=\frac{\sqrt{5}}{3},-2\sqrt{5}$
$\sqrt{5}\bigg(3\text{x}^2+\frac{25}{\sqrt{5}}\text{x}-\frac{10\sqrt{5}}{\sqrt{5}}\bigg)=0$
$\Rightarrow\sqrt{5}\big[3\text{x}^2+5\sqrt{5}\text{x}-10\big]=0$
$\big(\text{Dividing by}\sqrt{5}\big)$
$\Rightarrow\big[3\text{x}^2-\sqrt{5}\text{x}+6\sqrt{5}\text{x}-10\big]=0$
$\Rightarrow3\text{x}^2-\sqrt{5}\text{x}+6\sqrt{5}\text{x}-10=0$
$\Rightarrow\text{x}\big(3\text{x}-\sqrt{5}\big)+2\sqrt{5}\big(3\text{x}-\sqrt{5}\big)=0$
$\Rightarrow\big(3\text{x}-\sqrt{5}\big)\big(\text{x}+2\sqrt{5}\big)=0$
Either $3\text{x}-\sqrt{5}=0,$ then $3\text{x}=\sqrt{5}$
$\Rightarrow\text{x}=\frac{\sqrt{5}}{3}$
or $\text{x}+2\sqrt{5}=0,$ then $\text{x}=-2\sqrt{5}$
$\therefore\text{x}=\frac{\sqrt{5}}{3},-2\sqrt{5}$
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