Equation with radicals asked by Husni Elayyan Madarseh Madarseh in math group معلمو الرياضيات محافظة قلقيلية

Exercise:
Solve in $\mathbb{R}$, $x=\sqrt{21-\sqrt{21+x}}$
Solution: In order to crack this equation we will use zero trick
So that, $x=\sqrt{21+x-x-\sqrt{21+x}}=\sqrt{{{\left( \sqrt{21+x} \right)}^{2}}-\left( x+\sqrt{21+x} \right)}$
Put $w=\sqrt{21+x}$ then $x=\sqrt{{{w}^{2}}-\left( x+w \right)}\Leftrightarrow {{x}^{2}}={{w}^{2}}-\left( x+w \right)$
i.e $\left( x-w \right)\left( x+w \right)+\left( x+w \right)=0\Leftrightarrow \left( x+w \right)\left( x-w+1 \right)=0$
it follows that, $w=-x\,\,\,or\,\,\,x=w-1$
so that, $\sqrt{21+x}-1=x\Leftrightarrow {{\left( x+1 \right)}^{2}}=21+x\Leftrightarrow {{x}^{2}}+2x+1=21+x$
$\Rightarrow {{x}^{2}}+x-20=0\Leftrightarrow \left( x+5 \right)\left( x-4 \right)=0$
Thus the accepted roots for this equation is $x=4$