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:

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

Answer: C.

Hope it's clear.

(1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..?

I'm having confusion at this point...Please help me understand!
_________________

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

Answer: C.

Hope it's clear.

(1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..?

I'm having confusion at this point...Please help me understand!

(1) says that x=-16 OR x=4. The equation is \(x^2+12x-64=0\) (k=64) --> x=-16 OR x=4.
_________________

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

Answer: C.

Hope it's clear.

Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

Answer: C.

Hope it's clear.

Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan

Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0.

(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12x-k=0\), so we would have \((x+16)*(something)=0\). Thus x=-16 is one of the roots of the given quadratic equation.

Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Thus according to the above \(x_1+x_2=-16+x_2=\frac{-12}{1}\) --> \(x_2=4\).

So, we have that x is either -16 or 4. Not sufficient.

(2) x≠−16. Clearly insufficient.

(1)+(2) Since (2) says that x is NOT -16, then x=4. Sufficient.

Answer: C.

Hope it's clear.

Why A is not sufficient? X cannot be -16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan

Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0.

Hope this helps.

Thanks. I didnt know about factors of an expression.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...