Solve the Following Quadratic Equation by Factorization

Question:

\[ \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0 \]

Solution

Given:

\[ \sqrt{3}x^2-2\sqrt{2}x-2\sqrt{3}=0 \]

Product of the coefficient of \(x^2\) and the constant term:

\[ (\sqrt{3})(-2\sqrt{3})=-6 \]

We split the middle term \(-2\sqrt{2}x\) as \(-3\sqrt{2}x+\sqrt{2}x\):

\[ \sqrt{3}x^2-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0 \]

Taking common factors:

\[ x(\sqrt{3}x-3\sqrt{2}) +\sqrt{2}(\,x-\sqrt{6}\,) =0 \] \[ x(\sqrt{3}x-3\sqrt{2}) +\frac{\sqrt{2}}{\sqrt{3}}(\sqrt{3}x-3\sqrt{2}) =0 \] \[ (\sqrt{3}x-3\sqrt{2}) \left(x+\frac{\sqrt{6}}{3}\right)=0 \]

Therefore,

\[ \sqrt{3}x-3\sqrt{2}=0 \quad \text{or} \quad x+\frac{\sqrt{6}}{3}=0 \] \[ x=\frac{3\sqrt{2}}{\sqrt{3}} =\sqrt{6} \] \[ x=-\frac{\sqrt{6}}{3} \]

Final Answer

\[ \boxed{x=\sqrt{6} \text{ or } x=-\frac{\sqrt{6}}{3}} \]

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