Question
Solve the following quadratic equations by factorization:
$\sqrt{3}\text{x}^2-2\sqrt{2}\text{x}-2\sqrt{3}=0$
$\sqrt{3}\text{x}^2-2\sqrt{2}\text{x}-2\sqrt{3}=0$
✓
Answer
$\sqrt{3}\text{x}^2-2\sqrt{2}\text{x}-2\sqrt{3}=0$
Here, $\text{a}=\sqrt{3},\text{b}=-2\sqrt{2},\text{c}=-2\sqrt{3}$
$\text{D}=\text{b}^2-4\text{ac}$
$=\big(-2\sqrt{2}\big)^2-4\times\sqrt{3}\times\big(-2\sqrt{3}\big)$
$=8+24=32$
$\because\text{D}>0$
$\therefore$ The given equation has real roots,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}$
or $\frac{-\big(-2\sqrt{2}\big)\pm\sqrt{32}}{2\times\sqrt{3}}$
or $\text{x}=\frac{2\sqrt{2}\pm\sqrt{16\times2}}{2\sqrt{3}}$
$=\frac{2\sqrt{2}\pm4\sqrt{2}}{2\sqrt{3}}$
Either $\text{x}=\frac{2\sqrt{2}+4\sqrt{2}}{2\sqrt{3}}$ or $\text{x}=\frac{2\sqrt{2}-4\sqrt{2}}{2\sqrt{3}}$
$=\frac{6\sqrt{2}}{2\sqrt{3}}=\frac{3\sqrt{2}}{\sqrt{3}}$ or $=\frac{-2\sqrt{2}}{2\sqrt{3}}=\frac{-\sqrt{2}}{\sqrt{3}}$
$\Rightarrow\frac{3\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$ or $\frac{-\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\frac{3\sqrt{6}}{3}\text{ and }\frac{-\sqrt{6}}{3}$
$\Rightarrow\sqrt{6}\ \text{and}\ \frac{-\sqrt{6}}{3}$
$\therefore\text{x}=\sqrt{6}$ and $\frac{-\sqrt{6}}{3}$
Here, $\text{a}=\sqrt{3},\text{b}=-2\sqrt{2},\text{c}=-2\sqrt{3}$
$\text{D}=\text{b}^2-4\text{ac}$
$=\big(-2\sqrt{2}\big)^2-4\times\sqrt{3}\times\big(-2\sqrt{3}\big)$
$=8+24=32$
$\because\text{D}>0$
$\therefore$ The given equation has real roots,
$\text{x}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}$
or $\frac{-\big(-2\sqrt{2}\big)\pm\sqrt{32}}{2\times\sqrt{3}}$
or $\text{x}=\frac{2\sqrt{2}\pm\sqrt{16\times2}}{2\sqrt{3}}$
$=\frac{2\sqrt{2}\pm4\sqrt{2}}{2\sqrt{3}}$
Either $\text{x}=\frac{2\sqrt{2}+4\sqrt{2}}{2\sqrt{3}}$ or $\text{x}=\frac{2\sqrt{2}-4\sqrt{2}}{2\sqrt{3}}$
$=\frac{6\sqrt{2}}{2\sqrt{3}}=\frac{3\sqrt{2}}{\sqrt{3}}$ or $=\frac{-2\sqrt{2}}{2\sqrt{3}}=\frac{-\sqrt{2}}{\sqrt{3}}$
$\Rightarrow\frac{3\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$ or $\frac{-\sqrt{2}}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\frac{3\sqrt{6}}{3}\text{ and }\frac{-\sqrt{6}}{3}$
$\Rightarrow\sqrt{6}\ \text{and}\ \frac{-\sqrt{6}}{3}$
$\therefore\text{x}=\sqrt{6}$ and $\frac{-\sqrt{6}}{3}$
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