Question
Solve the following quadratic equation:
$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
✓
Answer
$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
Here, $6\sqrt3\times\sqrt3=6\times3=18$ and $9 \times 2 = 18$ & $9 + 2 = 11$
$\Rightarrow\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}^2+9\text{x}+2\text{x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+3\sqrt3\big)+2\big(\text{x}+3\sqrt{3}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{3}\big)\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\text{x}+3\sqrt3=0$ or $\sqrt3\text{x}+2=0$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2}{\sqrt3}$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2\times\sqrt3}{\sqrt3\times\sqrt3}=\frac{-2\sqrt3}{3}$
Hence, $-3\sqrt3$ and $\frac{-2\sqrt3}{3}$ are the roots of the given equation.
Here, $6\sqrt3\times\sqrt3=6\times3=18$ and $9 \times 2 = 18$ & $9 + 2 = 11$
$\Rightarrow\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}^2+9\text{x}+2\text{x}+6\sqrt3=0$
$\Rightarrow\sqrt3\text{x}\big(\text{x}+3\sqrt3\big)+2\big(\text{x}+3\sqrt{3}\big)$
$\Rightarrow\big(\text{x}+3\sqrt{3}\big)\big(\sqrt3\text{x}+2\big)=0$
$\Rightarrow\text{x}+3\sqrt3=0$ or $\sqrt3\text{x}+2=0$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2}{\sqrt3}$
$\Rightarrow\text{x}=-3\sqrt3$ or $\text{x}=\frac{-2\times\sqrt3}{\sqrt3\times\sqrt3}=\frac{-2\sqrt3}{3}$
Hence, $-3\sqrt3$ and $\frac{-2\sqrt3}{3}$ are the roots of the given equation.
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