<s>[INST] Solve the radical equation $ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$ 
Solve the following equation:
  $$ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$

I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated.
Can someone help me with this equation, please?
 [/INST] Assuming you want real roots, you can use this trick . . .

Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$.

Then $f$ is an odd function.

Also, $f$ is strictly increasing, hence $f$ is one-to-one.

Then, letting $u=2x+1$,
\begin{align*}
&x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0
\qquad\qquad\qquad\qquad\;\;
\\[4pt]
\iff\;&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\iff\;&f(x) + f(u)=0\\[4pt]
\iff\;&f(x) = -f(u)\\[4pt]
\iff\;&f(x) = f(-u)\qquad\text{[since $f$ is odd]}\\[4pt]
\iff\;&x = -u\qquad\qquad\;\;\text{[since $f$ is one-to-one]}\\[4pt]
\iff\;&x = -(2x+1)\\[4pt]
\iff\;&3x+1 = 0\\[4pt]
\iff\;&x = -{\small{\frac{1}{3}}}\\[4pt]
\end{align*}
As an alternative, using the same trick, you can get a factored form:
\begin{align*}
&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt]
\implies\;&x\sqrt{x^2+5}=-u\sqrt{u^2 + 5}\\[4pt]
\implies\;&x^2(x^2+5)=u^2(u^2 + 5)\\[4pt]
\implies\;&x^4+5x^2=u^4+5u^2\\[4pt]
\implies\;&(u^4-x^4)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2)+5(u^2-x^2)=0\\[4pt]
\implies\;&(u^2-x^2)(u^2+x^2+5)=0\\[4pt]
\implies\;&(u-x)(u+x)(u^2+x^2+5)=0\\[4pt]
\implies\;&(x+1)(3x+1)(5x^2+4x+6)=0\qquad\text{[replacing $u$ by $2x+1$]}
\end{align*}
The two candidate real roots, $x=-1,\;x=-{\large{\frac{1}{3}}}\;$need to be verified against the original equation since, when squaring both sides, extraneous real roots were potentially introduced. In this case, as it turns out, the candidate root $x=-{\large{\frac{1}{3}}}$ is ok, but the candidate root $x=-1$ fails, so is not an actual root.
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