Brent@GMATPrepNow wrote:What is the sum of all solutions to the equation x^(2/3) - x^(1/3) - 2 = 4?
A) -35
B) -19
C) 7
D) 19
E) 35
Source:
www.gmatprepnow.com
$$?\,\,\,:\,\,\,\,\,{\text{sum}}\,\,{\text{of}}\,\,{\text{roots}}\,\,{\text{of}}\,\,\,\,\root 3 \of {{x^2}} - \root 3 \of x - 6 = 0$$
$$y = \root 3 \of x \,\,\,\,\,\,\,\,\,\left[ {\,\root 3 \of {{x^2}} = {{\left( {\root 3 \of x } \right)}^2} = {y^2}\,} \right]\,$$
$${y^2} - y - 6 = 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits_{{\rm{product}}\, = \, - 6}^{{\rm{sum}}\, = \,1} \,\,\,\,\,\,\,\,\left\{ \matrix{
\,3 = {y_1} = \root 3 \of {{x_1}} \,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{cubing}}} \,\,\,\,\,{x_1} = {3^3} = 27 \hfill \cr
\, - 2 = {y_2} = \root 3 \of {{x_2}} \,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{cubing}}} \,\,\,\,\,{x_2} = {\left( { - 2} \right)^3} = - 8 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,?\,\, = \,\,27 + \left( { - 8} \right)\,\, = \,\,19$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.