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

Lennie Carroll 2021-10-01 Answered
Solve the equation.
\(\displaystyle{4}\sqrt{{{1}+{3}{x}}}+\sqrt{{{6}{x}+{3}}}=\sqrt{{-{6}{x}-{1}}}\)

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Expert Answer

Maciej Morrow
Answered 2021-10-02 Author has 13549 answers
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 \(\displaystyle{x}=−{\frac{{{1}}}{{{3}}}}\)
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