Simple Algebra Exercise

Exercise:

Solve in $\mathbb{R}$ , $\sqrt{1+x\sqrt{{{x}^{2}}+16}}=1+x$

Solution: Clearly $x=0$ is a root since $\sqrt{1+x\sqrt{{{x}^{2}}+16}}=\sqrt{1+0\sqrt{16}}=1=1+0$

 Squaring both sides to get $1+x\sqrt{{{x}^{2}}+16}={{\left( 1+x \right)}^{2}}\Leftrightarrow x\sqrt{{{x}^{2}}+16}={{\left( 1+x \right)}^{2}}-1$

Take $u=1+x$ $\Rightarrow \left( u-1 \right)\sqrt{{{\left( u-1 \right)}^{2}}+16}={{u}^{2}}-1=\left( u-1 \right)\left( u+1 \right)$ $\Rightarrow \sqrt{{{\left( u-1 \right)}^{2}}+16}=u+1$

Squaring both sides again to get ${{\left( u-1 \right)}^{2}}+16={{\left( u+1 \right)}^{2}}$ $\Rightarrow {{\left( u+1 \right)}^{2}}-{{\left( u-1 \right)}^{2}}=16$

$\Rightarrow \left( u+1-u+1 \right)\left( u+1+u-1 \right)=16$ $\Rightarrow 2\left( 2u \right)=16\Leftrightarrow u=4$ But $u=1+x\Rightarrow x=3$