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If z^2 - 4z > 5 then which of the following is always

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If z^2 - 4z > 5 then which of the following is always [#permalink]

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21 May 2008, 12:01
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If z^2 - 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

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21 May 2008, 12:30
Z < -1, Can anyone solve this algebraically?

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21 May 2008, 12:43
i get Z<-1

basically you get (z-5)(z+1)>0

for this to be positive.. z is either >5 which is not an answer..choice

or z<-1..

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Director
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21 May 2008, 12:56
bingo..forgot to move the 5 over...

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Manager
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20 Jun 2008, 11:12
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hey guys, here

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

z - 5 > 0 or z>5 ok...

similarly z+1>0 then shouldnt it be z>-1

moreover I wanted to know if we have -

Equation 1 (x+2) (x+4) < 0
solving -
-4<x<-2 is this correct

Eq 2 (x-2) (x-4) < 0
2<x<4 is this correct

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Current Student
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20 Jun 2008, 13:04
This is my thinking:

Either both z-5>0 and z+1>0 (positive times positive > 0) => z>5 and z>-1 => z>5
Or both z-5<0 and z+1<0 (negative times negative > 0) => z<5 and z<-1 => z<-1

Can somebody explain how z<-1 ALWAYS?

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Intern
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22 Jun 2008, 01:09
Hey guy's i still dont get it ....
should'nt the ans be z>-1 ????

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22 Jun 2008, 07:59
young_gun wrote:
If z^2 - 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

The original inequality is Z^2 - 4Z > 5

it means, Z(Z-4) >5 ;Z>5 or Z>9

Now if you see any inequality which covers -ve or +ve fraction (which is less than 1 or greater than -1) does not satisfy the equation.

only Z<-1 does not cover the 0<z<1 or -1<z<0 fractions hence its the answer.

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23 Jun 2008, 08:28
If z^2 - 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

z^2 - 4z > 5 ======= >

z^2 - 4z - 5 > 0 ==========>

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

z-5 > 0 or z+1>0 ====>

z>5 or z>-1

z > 5 is not there in any of the options..
so z> -1 is the answer...

correct me if i am wrong..

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Manager
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23 Jun 2008, 08:46
I'm lost ....

I'm reasoning like dipenambalia and getting z>-1. Is this wrong then? I don't understand how you get z<-1

Could someone pls explain? Thx

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23 Jun 2008, 09:02
Virflo wrote:
I'm lost ....

I'm reasoning like dipenambalia and getting z>-1. Is this wrong then? I don't understand how you get z<-1

Could someone pls explain? Thx

basically the stem become (z-5)(z-1)>0

if z>-1 then..say pick 2.. z-5=-3 z-1=1..-3*1 CANNOT BE >0

so Z<-1 i.e -2, -3 ....-infinity

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Director
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23 Jun 2008, 09:21
young_gun wrote:
If z^2 - 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

It's a REALLY BAD question. (z-5)*(z+1) > 0 -> either z > 5 or z < -1

There is a problem with logic. If z < -1 -> (z-5)*(z+1) > 0 but not the other way. for example, z = 6 satisfies (z-5)*(z+1) > 0. Does that mean that it satisfies z < -1 -> NO.

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CEO
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23 Jun 2008, 09:42
young_gun wrote:
If z^2 - 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

Inequalitys are usually one of my weak spots. Dealing with it algebraically that is.

My approach was to just use 0 for Z, which can be done for each except E.

Thus E it is.

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23 Jun 2008, 09:52
young_gun wrote:
If z^2 - 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

Note that z could be a large positive or a large negative number, so none of of the above is always true.

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23 Jun 2008, 09:56
GMATBLACKBELT wrote:
Inequalitys are usually one of my weak spots. Dealing with it algebraically that is.

My approach was to just use 0 for Z, which can be done for each except E.

Thus E it is.

And you would be correct if the question was right. But it is not. Answer is definitely not A, B, C or D, you are right with that. But nor is it E.

Hence z<-1 is not enough: it doesn't answer the question, since we cannot say "if z^2 - 4z > 5 then z is always less than -1" (try z=6 if you are not convinced).

So either this is not a GMAT question, or there is a mistake in it (in fact either way there is a mistake in it).

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23 Jun 2008, 11:04
If z^2 - 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

z^2 - 4z -5>0
(z+1) (z-5) > 0

z-5>0 or z+1 <0
Soln: z> 5 or z<-1
Since we don't see z > 5, z<-1 is the solution.
This is quadratic inequalities question, we can also solve it using graphs, the possible solution will be either z>5 or z<-1.

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24 Jun 2008, 09:55
maratikus wrote:
young_gun wrote:
If z^2 - 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

It's a REALLY BAD question. (z-5)*(z+1) > 0 -> either z > 5 or z < -1

There is a problem with logic. If z < -1 -> (z-5)*(z+1) > 0 but not the other way. for example, z = 6 satisfies (z-5)*(z+1) > 0. Does that mean that it satisfies z < -1 -> NO.

You can't deal with inequalities the same way you deal with equalities (in which you set the equation equal to zero and solve for the parts that make one or the other equal to zero). This is because, the value of the variable affects the other component of the inequality. A graphical representation might help:
Attachments

inequality.gif [ 9.73 KiB | Viewed 1029 times ]

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Director
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24 Jun 2008, 10:00
brokerbevo, you are wrong. Try z = 6. It satisfies the first inequality z^2-4z > 5 but doesn't satisfy option E : z < -1.

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24 Jun 2008, 10:02
No. (E) doesn't qualify to answer the question which was "which of the following is always true?"

What is always true, is that if z<-1, then z^2 - 4z > 5.

But if z^2 - 4z > 5, then we cannot say that always z<-1 (it could be that z>5)

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24 Jun 2008, 10:18
maratikus wrote:
brokerbevo, you are wrong. Try z = 6. It satisfies the first inequality z^2-4z > 5 but doesn't satisfy option E : z < -1.

Ahhh. I see what you are saying now. Yes, question is worded badly-- probably won't see this on GMAT. I spent all that time making that graph for nothing.
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Re: PS inequality   [#permalink] 24 Jun 2008, 10:18
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