fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 12)
$$x > 0\,\,,\,\,\,\,{x^{\sqrt x }} = {x^2}\,\,\,\left( * \right)$$
$$?\,\,:\,\,{\rm{sum}}\,\,{\rm{of}}\,\,{\rm{roots}}$$
$$x = 1\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{inspection}}} \,\,\,\,\left\{ \matrix{
\,\,{x^{\sqrt x }} = {1^{\sqrt 1 }} = 1 \hfill \cr
\,\,{x^2} = {1^2} = 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{first}}\,\,{\rm{root}}$$
$$0 < x < 1\,\,\,{\rm{or}}\,\,\,x > 1\,\,\,:\,\,\,\,\,$$
$$\left( * \right)\,\,\,\, \Rightarrow \,\,\,{x^{\sqrt x \, - \,2}} = 1 = {x^0}\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{base}}\,\, \ne \,\,0,1, - 1} \,\,\,\,\,\sqrt x - 2 = 0\,\,\,\, \Rightarrow \,\,\,\,\,x = 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{second}}\,\,{\rm{root}}$$
$$? = 1 + 4$$
The correct answer is (E).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.