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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] New post 13 Nov 2004, 08:16
If z^2-4z > 5, then which of the following is always true?

1. z>-5
2. z<5
3. z>-1
4. z<1
5. z<-1

The wording confused me and I dont think its right.
Lets see what you guys think of this question.
Please post your answers and then I'll reveal the OA.
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 [#permalink] New post 13 Nov 2004, 08:38
Solving this we have:

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

This is true only if:
1. (z-5) & (z+1) are both +ve... Z>5
2. (z-5) & (z+1) are both -ve... Z<-1

So the only tue solution is (5) Z < -1

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 [#permalink] New post 13 Nov 2004, 09:21
I got choice 3) as the answer....

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math [#permalink] New post 13 Nov 2004, 16:35
I don't get your answer, manger, how did you decide that your second choice was the right answer?

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 [#permalink] New post 14 Nov 2004, 06:13
Complex vision,
your ans is correct.
However, what confused me was the wording.
The question says 'which of the following must be true always'?

But as you showed, when x>5, the inequality holds true also.
So (this is becomng more of a logic problem now), if the inequality holdstrue, there is an alternative possibility (x>5) too. Hence we can not say that 'x<-1" always if you know the inequality holds!

Do you agree? I perferctly agree that out of the ans choices, what you have selected is the only right one. But the question doesnt sound right to me.
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 [#permalink] New post 14 Nov 2004, 06:15
Sorry, the gest post abve is actually mine.

I forgot to mention the source of the problem.
Its from one of the 800score tests.
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 [#permalink] New post 14 Nov 2004, 06:19
The wording is fine. If you have the choice b/w x>5 and x<-1, you will pick any answer that will be covered by those critical values. As a matter of fact, if the answer choices included x>10 or x<-10, they would also have been valid choices since those answers would have ALWAYS been true. Knowing that x<-1 is ALWAYS true does NOT mean that you are refuting the alternate answer x>5. It just means that you are accepting one answer as true nothwithstanding how true the alternate answer x>5 is.

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 [#permalink] New post 14 Nov 2004, 12:58
ComplexVision wrote:
Solving this we have:

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

This is true only if:
1. (z-5) & (z+1) are both +ve... Z>5
2. (z-5) & (z+1) are both -ve... Z<-1

So the only tue solution is (5) Z < -1


can anyone please explain what is the principle behind this??
thanks.

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 [#permalink] New post 14 Nov 2004, 15:40
Anonymous wrote:
Complex vision,
your ans is correct.
However, what confused me was the wording.
The question says 'which of the following must be true always'?

But as you showed, when x>5, the inequality holds true also.
So (this is becomng more of a logic problem now), if the inequality holdstrue, there is an alternative possibility (x>5) too. Hence we can not say that 'x<-1" always if you know the inequality holds!

Do you agree? I perferctly agree that out of the ans choices, what you have selected is the only right one. But the question doesnt sound right to me.


I totaly agree with you mbahope... The problem depends on a logic principle, which is:

In order to make (A OR B) true statement

then one of the following should apply:

1. A is true
2. B is true
3. A AND B is true


In our case, the statement is Z>5 OR Z<-1
So, if ONLY ONE of them applies, then that is sufficient to make the whole statement true

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 [#permalink] New post 15 Nov 2004, 06:12
Thanks everyone.
Like complex v said, for example some result could be true if

1. A is true
2. B is true.
3. A and B are true

However that does not mean that we can say, "If the result is true, then A MUST be true( because it could be caused by B being true)! ". We can say the otherway round, however. i.e. If B is true, the result has to be true.
My concern is the wording of the question should have made that clear. Instead the question just asked this "If the result is true, must B also be true?"
B doesnt HAVE to be true for the result to be true. The result could be true if A were true (and in this case A is not a subset of B).
You guys see what I'm trying to say..

And like i said in my prev post, i agree that out of the ans choices, the only right one was the E. But the WORDING of what has been asked "If the inequality is true, which of the following MUST be true" doesnt make sense.
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 [#permalink] New post 15 Nov 2004, 07:22
In this kind of "greater then" or "less then" problems, drawing a number lines helps. If you draw a number line and place the values, the answer becomes clear.

See if it works for you...

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 [#permalink] New post 15 Nov 2004, 07:33
Agree with Aspire2005. So you found the critical values which are 5 and -1. You have in total 3 segments. Plug in -2, 0 and 6 and you will find which answers work and which don't.

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 [#permalink] New post 15 Nov 2004, 08:35
I think thats a grest suggestion and couldnt have come from a better person!
I tried to solve the problem again using this suggestion and it became so much easier to eliminate the wrong choices.
Great tip, aspire2005 and congrats again for your awesome score!
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 [#permalink] New post 16 Nov 2004, 13:55
ComplexVision, can you explain how you got z < -1, when you solve the inequality you get (z-5)(z+1) > 0, so z > 5 or z > -1, where am I missing your point of z < -1
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 [#permalink] New post 17 Nov 2004, 12:50
aspire2005 wrote:
In this kind of "greater then" or "less then" problems, drawing a number lines helps. If you draw a number line and place the values, the answer becomes clear.

See if it works for you...


Hi, can you please show how does this work? I tried but didnt understand.

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 [#permalink] New post 17 Nov 2004, 18:29
ruhi - see the attachment. By solving the equation we got 2 numbers 5 and -1. Now, as Paul wrote, on the number line we get 3 segments - (a) numbers less then -1, (b) numbers between -1 and 5, (c) numbers greater then 5.

So, if you take one example from each and try out, you will figure out the answer :-D

P.S.: For the first time I used Microsoft Visio software to draw a number line :lol:

Attachments

Number.gif
Number.gif [ 2.52 KiB | Viewed 593 times ]


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 [#permalink] New post 17 Nov 2004, 22:44
Aspire,

Thanks a lot for that help. Though I could draw the numebr line, I couldnt understand what to do next. Thanks a lot :-D

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 [#permalink] New post 18 Nov 2004, 07:40
Paul wrote:
The wording is fine. If you have the choice b/w x>5 and x<-1, you will pick any answer that will be covered by those critical values. As a matter of fact, if the answer choices included x>10 or x<-10, they would also have been valid choices since those answers would have ALWAYS been true. Knowing that x<-1 is ALWAYS true does NOT mean that you are refuting the alternate answer x>5. It just means that you are accepting one answer as true nothwithstanding how true the alternate answer x>5 is.


great explanation paul. the number line aspire came up with is also very cool. Just beware of inequalities and quadratic equations/polynomials. What would be true if this were (z-5)(z+1)=0 is not true when it's (z-5)(z+1)>0. You've got to step in and use your logic a little more carefully. The test would love you to make that mistake, so of course they're going to put the option in to choose the wrong answer.
  [#permalink] 18 Nov 2004, 07:40
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