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# If z - 4z > 5 then which of the following is always true

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If z - 4z > 5 then which of the following is always true [#permalink]

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04 Nov 2006, 10:57
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If zÂ² - 4z > 5 then which of the following is always true

A) z > -5

B) z < 5

C) z > -1

D) z < 1

E) z < -1
SVP
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04 Nov 2006, 11:08
zÂ² - 4z > 5
<=> zÂ² - 4z - 5 > 0
<=> (z+1)(z-5) > 0

As zÂ² - 4z - 5 is the kind of a*x^2 + b*x + c where a is positive, thus (z-1)(z+5) > 0 is possible when z is not between the 2 roots : -1 and 5.

Hence, we know that z < -1 or z > 5

The question is a bit strange.... we cannot be sure here ... But as we have to choose, I take (E)
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06 Nov 2006, 09:22
Fig wrote:
zÂ² - 4z > 5
<=> zÂ² - 4z - 5 > 0
<=> (z+1)(z-5) > 0

As zÂ² - 4z - 5 is the kind of a*x^2 + b*x + c where a is positive, thus (z-1)(z+5) > 0 is possible when z is not between the 2 roots : -1 and 5.

Hence, we know that z < -1 or z > 5

The question is a bit strange.... we cannot be sure here ... But as we have to choose, I take (E)

I get:
z^2-4z-5>0
(z-5) (z+1)>0
z>5,z>-1

I do not understand the change in sign part. Pls explain.
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06 Nov 2006, 09:40
Sumithra wrote:
Fig wrote:
zÂ² - 4z > 5
<=> zÂ² - 4z - 5 > 0
<=> (z+1)(z-5) > 0

As zÂ² - 4z - 5 is the kind of a*x^2 + b*x + c where a is positive, thus (z-1)(z+5) > 0 is possible when z is not between the 2 roots : -1 and 5.

Hence, we know that z < -1 or z > 5

The question is a bit strange.... we cannot be sure here ... But as we have to choose, I take (E)

I get:
z^2-4z-5>0
(z-5) (z+1)>0
z>5,z>-1

I do not understand the change in sign part. Pls explain.

We apply this rule :
> f(x) = a*x^2 + b*x + c where a is positive, f(x) > 0 when x is not between its roots if these roots exist.
> f(x) = a*x^2 + b*x + c where a is negative, f(x) > 0 when x is between its roots if these roots exist.

If we represent f(z)=z^2-4z-5 in XY plan, we obtain the fig1.

I also add u the representation of g(z)=-z^2-4z+5 to have the case with a <0 in the fig2.
Attachments

Fig1_a-postive.jpg [ 17.01 KiB | Viewed 610 times ]

Fig2_a-negative.jpg [ 16.88 KiB | Viewed 609 times ]

Senior Manager
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06 Nov 2006, 09:41
solving,
we get

(z-5)(Z+1)>0

this holds only if z>5 or z<-1

hence z<-1..choice E

what's the oa???

yogeshsheth wrote:
If zÂ² - 4z > 5 then which of the following is always true

A) z > -5

B) z < 5

C) z > -1

D) z < 1

E) z < -1
SVP
Joined: 01 May 2006
Posts: 1796
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06 Nov 2006, 09:43
The problem is that we cannot be sure that z < -1.... The case z > 5 remains .... Strange question: a kaplan one ? :p

AK47 wrote:
solving,
we get

(z-5)(Z+1)>0

this holds only if z>5 or z<-1

hence z<-1..choice E

what's the oa???

yogeshsheth wrote:
If zÂ² - 4z > 5 then which of the following is always true

A) z > -5

B) z < 5

C) z > -1

D) z < 1

E) z < -1
SVP
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Posts: 1796
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06 Nov 2006, 09:52
Sumithra wrote:
Thank you Fig.

U are welcome .... I suggest u to have a look perhaps on this thread.... To complete the idea of sign, I represented a table of sign

http://www.gmatclub.com/phpbb/viewtopic.php?t=36468
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14 Nov 2006, 11:16
I get A. z^2-4z-5>0, roots are therefore -1 and +5. So Z is always >-5, since it's either -1 or 5...odd question though.
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14 Nov 2006, 11:28
The only right answer must be E.

We know that the other ones can't be, although E is not perfect as was said before.
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14 Nov 2006, 21:02
I got E for this one. My question is why is it this question an odd one ?? examining other choices, only x<-1 is one of the two solution. the other solution which is x>5 is not there and x> -5 or x>-1 range is too wide to make x always true.
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14 Nov 2006, 21:58
z^2 - 4z - 5 > 0
(z-5)(z+1) > 0

valid ranges:
z > 5, z < -1

E is always true.
14 Nov 2006, 21:58
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