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

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GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935

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10 Mar 2019, 09:37
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Difficulty:

55% (hard)

Question Stats:

60% (02:18) correct 40% (01:55) wrong based on 50 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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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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GMATH Teacher
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Re: Is (x^2-7x+12)(x^2-5x-14)>0?  [#permalink]

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

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