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z² - 4z > 5

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by figs » Mon May 25, 2009 4:33 am
If you have a look to all the answers except E, you can see that all the answers include the number 0 (except E).
and 0-0=0 and it's not bigger than 5.
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by Giorgio » Mon May 25, 2009 4:49 am
It looks like this:

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

There are 2 possible outcomes for this to be true.

1. X-5>0 X> 5
x+1>0 x>-1 So X must be greater than 5 , as it includes both intervals.

2. X-5<0 X<5
x+1<0 x<-1 X must be less than -1 to include both intervals....

So if you choose from your answer choices the correct answer is E. X must be less than -1.... another possible answer could be X >5 ... but it is not included in answers.

Hope it helps.
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by iikarthik » Mon May 25, 2009 5:42 am
Thanks for your replies.

but i tried to solve like this:

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

equating (z-5)>0 we get z>5


equating (z+1)>0 we get z>-1

so i chose C.

I want to know what went wrong in my calculation.

need your assistance :roll:
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by avanishjoshi » Mon May 25, 2009 7:30 am
You need to look at the common solution that will satisfy the inequality.

so if (Z-5)(Z+1) > 0,. then the area covering Z - 5 >0 and Z+1 > 0 or Z- 5 < 0 and Z+1<0 will satisfy the inequality.

Now if we choose Z - 5 > 0 and Z + 1 > 0 then the common area that satisfies this equation is z> 5 as represented below on the number line.

------|{-----------|---------------|{------------
-1 0 5
The common area that satisfies the equation is Z > 5
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Re: z² - 4z > 5

by maihuna » Mon May 25, 2009 8:41 am
iikarthik wrote:If z² - 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(z-4)>5
z(z-4)-5>0

-------------------|---------------|-----------
z=-1 no, -2 ok 0 for z=2, -4 4 for z=5, 5no, for z=6 12 yes

so for z<-1 z(z-4) will be always greter than 5 E is indeed corect

for z>-1 means somewhere right side z>0 which is incorrect
Charged up again to beat the beast :)
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by cramya » Mon May 25, 2009 8:47 am
z^2-4z-5>0

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

Case 1: z>-1 and z<5

Case 2: z<-1 and z<5

Let take z=-1/2 using case1 the condition will fail so z<-1 using case2 must always be true
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by Svedankae » Tue May 26, 2009 7:29 am
what is the source of this question?
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