Is (x^2-7x+12)(x^2-5x-14)0?

Question

GMATH practice exercise (Quant Class 14) Is (x^2-7x+12)(x^2-5x-14)>0 ? (1) x>0 (2) 3<|x|<4

CAtluri · Answer

Solving the equation gives 
(x-3)(x-4)(x-7)(X+2)>0

S1 gives nothing
S2 Try with -3.5 and 3.5 as x. Both satisfy the equation

So B

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fskilnik · Answer

\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.

Is (x^2-7x+12)(x^2-5x-14)>0?

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