Nice exercise posted by Karam Makhlof Knama in their facebook wall

Exercise:

Solve in $\mathbb{R}$, $\sqrt{x+4+2\sqrt{x+3}}+3\sqrt{x+4-2\sqrt{x+3}}=\left( x+2 \right)\sqrt{x+3}$

Solution: we have \[x+4+2\sqrt{x+3}=x+3+1+2\left( 1 \right)\sqrt{x+3}={{\left( \sqrt{x+3}+1 \right)}^{2}}\] and

$x+4-2\sqrt{x+3}=x+3+1-2\left( 1 \right)\sqrt{x+3}={{\left( \sqrt{x+3}-1 \right)}^{2}}$ ,then

$\sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\sqrt{{{\left( \sqrt{x+3}-1 \right)}^{2}}}=\left( x+2 \right)\sqrt{x+3}$

However, $\frac{1}{\sqrt{x+3}+1}\times \frac{\sqrt{x+3}-1}{\sqrt{x+3}-1}=\frac{\sqrt{x+3}-1}{x+3-1}=\frac{\sqrt{x+3}-1}{x+2}$

$\Rightarrow \left( \sqrt{x+3}+1 \right)\left( \sqrt{x+3}-1 \right)=x+2$

$\sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\sqrt{{{\left( \frac{x+2}{\sqrt{x+3}+1} \right)}^{2}}}=\left( x+2 \right)\sqrt{x+3}$

$\Rightarrow \sqrt{{{\left( \sqrt{x+3}+1 \right)}^{2}}}+3\left( x+2 \right)\sqrt{\frac{1}{{{\left( \sqrt{x+3}+1 \right)}^{2}}}}-\left( x+2 \right)\sqrt{x+3}=0$

Put $y=\sqrt{x+3}>0$ then ${{y}^{2}}-3=x$

$\Rightarrow \sqrt{{{\left( y+1 \right)}^{2}}}+3\left( {{y}^{2}}-3+2 \right)\frac{1}{\sqrt{{{\left( y+1 \right)}^{2}}}}-\left( {{y}^{2}}-3+2 \right)y=0$

$\Rightarrow y+1+\frac{3{{y}^{2}}-3}{y+1}-{{y}^{3}}+y=0$ multiply both sides by $y+1$ to get

$\Rightarrow {{\left( y+1 \right)}^{2}}+3\left( {{y}^{2}}-1 \right)-y\left( {{y}^{2}}-1 \right)\left( y+1 \right)=0\Leftrightarrow \left( y+1 \right)\left[ y+1+3\left( y-1 \right)-y\left( {{y}^{2}}-1 \right) \right]=0$

$\Rightarrow y=-1$ rejected or $y+1+3y-3-{{y}^{3}}+y=0\Leftrightarrow -{{y}^{3}}+5y-2=0$

\[\Rightarrow {{y}^{3}}-y-4y+2{{y}^{2}}-2{{y}^{2}}+2=0\]

$\Rightarrow {{y}^{3}}-y+2{{y}^{2}}-4y-2{{y}^{2}}+2=y\left( {{y}^{2}}+2y-1 \right)-2\left( {{y}^{2}}+2y-1 \right)=0$

$\Rightarrow \left( {{y}^{2}}+2y-1 \right)\left( y-2 \right)=0\Leftrightarrow {{y}^{2}}+2y-1=0\,\,\,or\,\,\,y=2$

But ${{y}^{2}}+2y+1-1-1={{\left( y+1 \right)}^{2}}=2\Leftrightarrow y=-1\pm \sqrt{2}$ but $y=-1-\sqrt{2}<0$ rejected

If $y=2\Leftrightarrow x+3=4\Leftrightarrow$ $x=1$. However, $y=-1\pm \sqrt{2}$ with

$x=-2\sqrt{2}$ is rejected after substation in the original equation