Solve the equation. 4\sqrt{1+3x}+\sqrt{6x+3}=\sqrt{-6x-1}

Solve the equation.
$$\displaystyle{4}\sqrt{{{1}+{3}{x}}}+\sqrt{{{6}{x}+{3}}}=\sqrt{{-{6}{x}-{1}}}$$

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Step 1
Given:
$$\displaystyle{4}\sqrt{{{1}+{3}{x}}}+\sqrt{{{6}{x}+{3}}}=\sqrt{{-{6}{x}-{1}}}$$
Step 2
The objective is to solve the equation
$$\displaystyle{4}\sqrt{{{1}+{3}{x}}}+\sqrt{{{6}{x}+{3}}}=\sqrt{{-{6}{x}-{1}}}$$
On removing the roots
$$\displaystyle{1152}{x}^{{{2}}}+{960}{x}+{192}={3600}{x}^{{{2}}}+{2400}{x}+{400}$$
On solving,
$$\displaystyle{1152}{x}^{{{2}}}+{960}{x}+{192}={3600}{x}^{{{2}}}+{2400}{x}+{400}$$
$$\displaystyle{x}=-{\frac{{{1}}}{{{3}}}},{x}=-{\frac{{{13}}}{{{51}}}}$$
Step 3
On verifying the solution
$$\displaystyle{x}=-{\frac{{{1}}}{{{3}}}}$$ it is true
$$\displaystyle{x}=-{\frac{{{13}}}{{{51}}}}$$ it is false
Therefore the solution is $$\displaystyle{x}=−{\frac{{{1}}}{{{3}}}}$$