Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 )

Fig can you elaborate on the logic of your answer ( Z level students teaching style plz?

thanks in advance

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

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..
_________________

ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED

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

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.

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 747 times ]

Signs-Table.jpg [ 14.17 KiB | Viewed 744 times ]

Last edited by Fig on 11 Oct 2006, 23:49, edited 2 times in total.

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.