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 350,000 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:

Re: QR:DS 62 Algebra Equation [#permalink]
21 Feb 2011, 05:40

2

This post received KUDOS

Expert's post

Baten80 wrote:

In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x-3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0

You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Re: QR:DS 62 Algebra Equation [#permalink]
21 Feb 2011, 05:56

Bunuel wrote:

Baten80 wrote:

In the equation x^2+bx+12=0, is variable and b is a constant. What is the value of b? (1) x-3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0

You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.

Agree!!! This is much easier and way faster than the Viete's formula (I didn't know that's what it's called). Double thanks!! _________________

Re: QR:DS 62 Algebra Equation [#permalink]
21 Feb 2011, 05:59

Bunuel wrote:

Baten80 wrote:

In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b? (1) x-3 is a factor of x^2+bx+12. (2) 4 is a factor of x^2+bx+12=0

You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.

First post was wrong. This the correct post. But I think answer is same. _________________

Re: In the equation ... [#permalink]
28 Feb 2011, 01:50

For equation x^2 + bx + 12 = 0, we need to find value of b

1) says x-3 is a factor of x^2+bx+12, or x=3 is one of the roots of this quadratic equation, so 3^2+3b+12 = 0 or 21+3b = 0 or b = -7. Sufficient 2) says 4 is one of the roots so 4^2+4b+12 = 0 or 28+4b = 0 or b = -7. Again sufficient, so D

if x-a is a factor of any polynomial in x, then at x=a, the value of such a polynomial would be 0 or in other words x = a is one of the roots of such a polynomial

Re: In the equation x^2+bx+12=0, x is a variable and b is a [#permalink]
03 Oct 2013, 04:49

D... just find factors, already x-3, x-4 is another factor to get 12 and -3x-4x= -7, this is possible becz x2+ bx +12 = 0, here b = +/-, to proof equation only is b= -7

Re: QR:DS 62 Algebra Equation [#permalink]
24 Oct 2013, 04:53

Bunuel wrote:

Baten80 wrote:

In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x-3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0

You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.

Im not really following your first point here.

How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation?

Re: QR:DS 62 Algebra Equation [#permalink]
24 Oct 2013, 06:49

Expert's post

waltiebikkiebal wrote:

Bunuel wrote:

Baten80 wrote:

In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x-3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0

You don't really need Viete's formula for the roots of a quadratic equation.

(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.

(2) 4 is a root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.

Answer: D.

Im not really following your first point here.

How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation?

If (x+2)(x+6)=0, then x-3 is NOT a factor of the quadratics. The roots of (x+2)(x+6)=0 are x=-3 and x=-6.

We are told that x-3 is a factor of x^2+bx+12, thus (x-3)(x-k)=0, thus x=3 is one of the roots.

hey guys, A metallurgist but currently working in a NGO and have scheduled my GMAT in December for second round .....u know. I read some but valuable blogs on this...

Today, 1st year Rotman students had a great simulation event hosted by Scotiabank, one of Canada’s best and largest banks. Attended by entire Rotman 1st year students, the...

Nope. I never learned finance ever in my life until I came to Rotman. This is why I got really frustrated when this term started because I was certain...