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# if x(x-5)(x+2)=0, is x negative? 1)x^2 - 7x >= 0 2) x^2 -

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Director
Joined: 02 Mar 2006
Posts: 582
Location: France
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How can it be C?

I find B too!

x=0 or 5 or -2

5 is the only solution for both equations, so x>0 and B sufficient.
Intern
Joined: 07 Oct 2005
Posts: 3
Followers: 0

Kudos [?]: 2 [0], given: 0

i think they arrived the answer this way (i cannot think of any other means)

if x(x-5)(x+2)=0, is x negative?

x^2 - 7x >= 0
x^2 - 2x - 15>=0

(subtracting)
-----------------------------
5x +15 >= 0

SVP
Joined: 03 Jan 2005
Posts: 2250
Followers: 13

Kudos [?]: 216 [0], given: 0

Ok, a couple things here.

First, I agree that C is not the right answer and I agree with D.

Second is for srknori. You can not subtract two inequalities when their signs are toward the same direction.
For example, if you have x>5, and also y>3, you cannot subtract the two inequalities and arrive x-y>2. It's clear that it may not be the case. Say x is 6 and y is also 6, then x-y=0, certainly not greater than 2.
You can only do subtraction when the signs for two inequalities are toward different direction. For example, if we know x>5 and y<3, then we know for sure that x-y>2.

Raghavender wrote:

thus, (x-5)(x+3) >=0
<=> x >= 5 or x <= -3

does x+3 >=0, not mean tht x>= -3( when u subtract 3 on both sides...)
please clear this doubt for me.....

When you have something like this, there is a relatively straight forward way to find the solution set. Look at this example:
(x-a)(y-b)<0
Remember for the product to be negative one has to be positive and the other has to be negative. So simply do two cases: x-a>0, y-b<0; and x-a<0, y-b>0. The result would be all possible solutions.
In our case, we have (x-5)(x+3)>=0. For the product to be non-negative, either both have to be non-negative, or both have to be non-positive. (Note here I use non-negative instead of positive because we need to include the case of zero.) So you list two options: x-5>=0, x+3>=0 and x-5<=0, x+3<=0. Now the solution set for the first set is x>=5, and the solution set for the second set is x<=-3.

The sticky math basic principle thread has some discussions about inequalites. I would recommend people who wants to refresh on this topic go spend some time to read it. It may help.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Last edited by HongHu on 13 Oct 2006, 04:37, edited 1 time in total.
Manager
Status: Post MBA, working in the area of Development Finance
Joined: 09 Oct 2006
Posts: 170
Location: Africa
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Kudos [?]: 3 [0], given: 1

Re: DS is x negative? (Q0407) [#permalink]  12 Oct 2006, 21:36
imaru wrote:
if x(x-5)(x+2)=0, is x negative?

1)x^2 - 7x >= 0
2) x^2 - 2x - 15>=0

1) x^2 - 7x >= 0 Implies that x>=7 or x<=0
Combine this with the question and for value of x=0, x(x-5)(x+2)=0 so the value of x being 0 makes it positive (or rather non-negative).
So statement 1) is sufficient

As has been explained by others, for the value of x = 5 as derived from st. 2, x(x-5)(x+2)=0 so st. 2 is also sufficient.!!!

Thus both the statements appear to be independently sufficient.

Am I at different (and wrong!) plane than others?

Will someone explain pls.?
SVP
Joined: 03 Jan 2005
Posts: 2250
Followers: 13

Kudos [?]: 216 [0], given: 0

You are right. For (1) we can identify x=0 and thus the stem equation=0 ie. not negative. So it is also sufficient.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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