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

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09 Oct 2006, 08:42
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if x(x-5)(x+2)=0, is x negative?

1)x^2 - 7x >= 0
2) x^2 - 2x - 15>=0
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09 Oct 2006, 08:55
B it is
(1) is insuff
Because x^2-7x>=0 means that x>=7 or x<=0
x(x-5)(x+2)=0 so x could be 0;5;-2
(1) only we get x= 0 (not negative) or -2 (negative)
So B is correct (x can get only one value 5 )
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09 Oct 2006, 10:06
if x(x-5)(x+2)=0, is x negative?

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

from stem

x is either = 0 or 5 or -2 question asks is x = -2
substitute in one

f(0) = could be , f(5) = wrong f(-2) = right.........insuff

from two

f(0) = wrong , f(5) = could be f(-2) = wrong....insuff

BUT as far as i know GMAT doesnt give the two statment with coontradiction ( case of f(-2))............. i doubt this is a gmat question.
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09 Oct 2006, 13:01
(B) as well,

Stat1:
x^2 - 7x >= 0
<=> x*(x-7) >=0
<=> x <= 0 or x >= 7

Since x(x-5)(x+2)=0, hence x = 0 or x = -2

INSUFF

Stat2:
x^2 - 2x - 15>=0
<=> (x-5)(x+3) >=0
<=> x >= 5 or x <= -3

Since x(x-5)(x+2)=0, hence x is forced to be equal to 5.

SUFF
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09 Oct 2006, 13:43
Fig can you elaborate on the logic of your answer ( Z level students teaching style plz?

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09 Oct 2006, 14:03
yezz wrote:
Fig can you elaborate on the logic of your answer ( Z level students teaching style plz?

At least, I can try

Since x(x-5)(x+2)=0, then x could be -2, 0 or 5.

If we have a close look on these possible solutions, we can observe that 1 is negative, 1 is without sign and 1 is positive.

To answer the problem quesion :'is x negative?', we could have these groups of solutions:
> groupe 1 : x = 5 (Not neg)
> groupe 2 : x = 5 or x = 0 (Not neg)
> groupe 3 : x = 0 (Not neg)
> groupe 4 : x = -2 (Neg)

Now, we could look at the statments. These statments have normally to give us 1 of the above group in order to respond all answers except (E).

Stat1: x^2 - 7x >= 0

Thus, x*(x-7) >=0
<=> x <= 0 or x >= 7

The bold inequality shows us that x is negative or 0. It's not match to 1 of the 4 groups definied to answer the question. -2 is negative but 0 is neither negative nor positive.
INSUFF

Stat2: x^2 - 2x - 15>=0

Thus, (x-5)(x+3) >=0
<=> x >= 5 or x <= -3
Bingo, -2 and 0 are out the possible values of x while x=5 (groupe 1) is contained in the bold inequality.
SUFF

Hope this help
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09 Oct 2006, 14:25

man i need to have some more rest........you rock my mate. thanks a million
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09 Oct 2006, 20:02
Hey ,

Getting bk to the inequality basics,...i'm awfully weak in inequalities...

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.....it wud help me a lot..coz inequality probs always pull me down..
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ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED

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10 Oct 2006, 10:46
Raghavender wrote:
Hey ,

Getting bk to the inequality basics,...i'm awfully weak in inequalities...

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.....it wud help me a lot..coz inequality probs always pull me down..

Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of :
o a (1 here) when x is not between the 2 roots, thus > 0
o -a (-1) when x is between the 2 roots, thus < 0

By the way, we have so:
(x-5)(x+3) >=0
<=> x >= 5 or x <= -3
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11 Oct 2006, 07:38
the OA is C...very weird...
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11 Oct 2006, 07:42
imaru wrote:
the OA is C...very weird...

What is the source of the question?
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11 Oct 2006, 07:47
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11 Oct 2006, 09:59
Fig wrote:
Raghavender wrote:
Hey ,

Getting bk to the inequality basics,...i'm awfully weak in inequalities...

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.....it wud help me a lot..coz inequality probs always pull me down..

Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of :
o a (1 here) when x is not between the 2 roots, thus > 0
o -a (-1) when x is between the 2 roots, thus < 0

By the way, we have so:
(x-5)(x+3) >=0
<=> x >= 5 or x <= -3

Fig - Thanks for the explanation. I kinda get it but am still a bit confused. How do you get x<=-3 ...when I subtarct three from x+3>=0 I get x>=-3. Thanks very much in advance.
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11 Oct 2006, 11:11
Matrix02 wrote:
Fig wrote:
Raghavender wrote:
Hey ,

Getting bk to the inequality basics,...i'm awfully weak in inequalities...

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.....it wud help me a lot..coz inequality probs always pull me down..

Actually, the equation (x-5)(x+3) = x^2 - 2x - 15 = a*x^2 + b*x + c has the sign of :
o a (1 here) when x is not between the 2 roots, thus > 0
o -a (-1) when x is between the 2 roots, thus < 0

By the way, we have so:
(x-5)(x+3) >=0
<=> x >= 5 or x <= -3

Fig - Thanks for the explanation. I kinda get it but am still a bit confused. How do you get x<=-3 ...when I subtarct three from x+3>=0 I get x>=-3. Thanks very much in advance.

Well, I can try

To seach (x-5)(x+3) >=0 is similar to study the sign of the function f(x) = (x-5)(x+3).

A great tool that Maths gives us is the table of signs. Once a fonction is factorized in "famous" forms of basic functions, such as a*x+b, we decompose the function and use the properties of the multiplication to determine what values of x give a positive sign and what values of x give a negative sign to the studied function f(x).

This table of signs is fast to draw and brings to the solution

I attached u the resulting table. To find the result signs for f(x), we multiply vertically + and - as they are +1 and -1.

So, when x <= -3 : sign(f(x)) = sign( (x-5)(x+3) ) = (-1)*(-1) = +1

I also join u the draw of the function f(x). U will visually remark that f(x) is positive on values of x not between the 2 roots.

This result is linked to my former post. The sign of a*x^2 + b*x + c
Attachments

Graph.jpg [ 31.16 KiB | Viewed 778 times ]

Signs-Table.jpg [ 14.17 KiB | Viewed 775 times ]

Last edited by Fig on 12 Oct 2006, 00:49, edited 2 times in total.
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11 Oct 2006, 11:17
Hell no...........Fig am no engneering expert like you are ....and obviously most of the people on this forum.

Man if this is Gmat I d rather kill myself ...

Cant we have a premetive down to earth explanation for prople like me ............I know u can Plzzzzzzzz.
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11 Oct 2006, 11:39
yezz wrote:
Hell no...........Fig am no engneering expert like you are ....and obviously most of the people on this forum.

Man if this is Gmat I d rather kill myself ...

Cant we have a premetive down to earth explanation for prople like me ............I know u can Plzzzzzzzz.

Well Sorry ... the table of signs is my best way to explain (x+3)(x-5) without possible confusions .

The famous result to remember is :
sign(a*x^2+b*x+c) = sign(-a) between the roots.

In this DS, we apply it 2 times
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11 Oct 2006, 11:45
Thanks so much FIG for the great explanation. I need to take some time to let it sink in and try some more problems. Great graphs..I really appreciate it.
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11 Oct 2006, 11:47
Ok Fig then ........on Gday I will guess
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11 Oct 2006, 12:06
U are welcome Matrix02

yezz... I believe that u will not guess but rather get it right
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11 Oct 2006, 12:10
This is the spirit..........man you are the man
11 Oct 2006, 12:10

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