fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 14)
Is (x^2-7x+12)(x^2-5x-14) > 0 ?
(1) x > 0
(2) 3 < |x| < 4
$$\left( {x - 3} \right)\left( {x - 4} \right)\left( {x + 2} \right)\left( {x - 7} \right)\,\,\mathop > \limits^? \,\,0$$
$$\left( 1 \right)\,\,x > 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 3\,\,\,\, \Rightarrow \,\,\,\left( {x - 3} \right)\left( {x - 4} \right)\left( {x + 2} \right)\left( {x - 7} \right) = 0\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = 8\,\,\,\, \Rightarrow \,\,\,\underbrace {\left( {x - 3} \right)}_{ > \,\,0}\underbrace {\left( {x - 4} \right)}_{ > \,\,0}\underbrace {\left( {x + 2} \right)}_{ > \,\,0}\underbrace {\left( {x - 7} \right)}_{ > \,\,0} > 0\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,$$
$$\left( 2 \right)\,\,3 < \left| x \right| < 4\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \matrix{
\,\,3 < x < 4\,\,\, \Rightarrow \,\,\,\underbrace {\left( {x - 3} \right)}_{ > \,\,0}\underbrace {\left( {x - 4} \right)}_{ < \,\,0}\underbrace {\left( {x + 2} \right)}_{ > \,\,0}\underbrace {\left( {x - 7} \right)}_{ < \,\,0}\,\,\,\, > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,\,\,\,{\rm{or}} \hfill \cr
\, - 4 < x < - 3\,\,\, \Rightarrow \,\,\,\underbrace {\left( {x - 3} \right)}_{ < \,\,0}\underbrace {\left( {x - 4} \right)}_{ < \,\,0}\underbrace {\left( {x + 2} \right)}_{ < \,\,0}\underbrace {\left( {x - 7} \right)}_{ < \,\,0}\,\,\,\, > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
