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

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Is (x^2-7x+12)(x^2-5x-14)>0?  [#permalink]

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New post 10 Mar 2019, 09:37
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A
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Difficulty:

  55% (hard)

Question Stats:

61% (02:35) correct 39% (01:53) wrong based on 38 sessions

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GMATH practice exercise (Quant Class 14)

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

(1) \(x>0\)
(2) \(3<|x|<4\)

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Re: Is (x^2-7x+12)(x^2-5x-14)>0?  [#permalink]

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New post 10 Mar 2019, 09:54
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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Re: Is (x^2-7x+12)(x^2-5x-14)>0?  [#permalink]

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New post 10 Mar 2019, 17:01
fskilnik 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.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Re: Is (x^2-7x+12)(x^2-5x-14)>0?   [#permalink] 10 Mar 2019, 17:01
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