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:

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

26 Jul 2014, 01:11

3

This post received KUDOS

2

This post was BOOKMARKED

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

_________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

29 Jul 2014, 20:55

2

This post received KUDOS

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

30 Jul 2014, 04:41

hubahuba wrote:

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

You are Welcome. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

18 Aug 2014, 13:48

VenoMftw wrote:

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

When you reach this point, how do you know that solving will give vales that won't be integers? Did you actually try to solve or do you have a shortcut?

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

19 Aug 2014, 17:34

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

I couldn't understand this line k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values. Where does the question say that the sum and product of the roots are integers.
_________________

I couldn't understand this line k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values. Where does the question say that the sum and product of the roots are integers.

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

23 Aug 2015, 22:25

Hello from the GMAT Club BumpBot!

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.
_________________

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

27 Aug 2016, 03:28

Hello from the GMAT Club BumpBot!

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.
_________________

If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

27 Oct 2016, 06:36

Hey, basically I came to the correct answer B but in other way:

(1) x1*x2=6=k/2 k=12 2x^2+hx+12=0

Basicaly, roots might be integers or non-integers so insufficient.

(2) h*k=132 (I) x1+x2=-h/2 (II) x1*x2=k/2 (III)

then I multiply (II) * (III) and get x1*x2*(x1+x2)=-k*h/4=-132/4=-33

So, if we asssume that x1 and x2 integers, what we should conclude is INTEGER1*INTEGER2*INTEGER3=-33 (which is possible only for triplet 1/3/11 with one/all of them negative - clearly among 1/3/11 there are no pairs to satisfy a+b=c =>> therefore, x1 OR x2 should be non-integer). So (2) is sufficient.

Not sure is it correct reasoning, but the answer is OK.

If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

01 Nov 2016, 00:21

VeritasPrepKarishma wrote:

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

It is not given that sum and product need to be integers. The logic of this question is a little more twisted:

Question: Are both roots positive?

Answer: Yes/No Or May be If we answer with a certain Yes or a certain No, our stmnt is sufficient. If we answer with May be (sometimes Yes, sometimes No), our statement is not sufficient.

2x^2 + hx + k = 0 Sum of roots = -h/2 Product of roots = k/2

Let's find all those cases in which we can answer with 'Yes'

Stmnt 1: The product of the roots is 6. k/2 = 6 so k = 12

The roots could be -2, -3 such that the equation would be x^2 + 5x + 6 = 0 or 2x^2 + 10x + 12 = 0 The roots could be no integer for any random value of h. So we have no definite answer from this statement. Not sufficient.

Stmnt 2: kh = 132 Let's see whether given this information, we can find both integer roots. kh = 132 = 2^2 * 3 * 11 If both roots are integers, sum and product of roots must be integers (k/2 and h/2 must be integers) So k and h must be even. So give a 2 to each of k and h and split 3 and 11 in two ways (no need to worry about negative sign): k and h could be 2 and 66 such that product and sum will be 1 and 33 - Can the product/sum of two integers be 33 while sum/product is 1? No. k and h could be 6 and 22 such that product and sum will be 3 and 11 - Can the product/sum of two integers be 11 while sum/product is 3? No. So it is not possible to have both integers. We can answer with a definite 'No'.

Re: If h and k are both positive integers, are both roots of the equation [#permalink]

Show Tags

05 Nov 2016, 05:43

1

This post was BOOKMARKED

hubahuba wrote:

VenoMftw wrote:

hubahuba wrote:

If h and k are both positive integers, are both roots of the equation 2 (x^2) + hx + k = 0 integers?

1. The product of the roots is 6 2. kh = 132

Source: Knewton

Can anyone help provide a simpler solution on this? It shows up as a challeneg question under Knewton.

Thanks.

Let the roots of the equation be P and Q. p+q = -h/2--(i) & pq = k/2 --(ii).

Statement 1 -- given pq= 6 i.e k = 12. From this we cant determine the exact solution.

Statement 2 -- Given kh =132 132 = 132 * 1 = 66 * 2 = 44 * 3 = 33 * 4 = 22 * 6 = 11 * 12 Thus these are the possible values k or h can take since given they are positive integers.

k or h cannot be odd. Since from equation -(i) and (ii) , if they are odd then h/2 or k/2 will give decimal values.

Thus k or h can take (66,2) (22, 6) if we try to solve the equations (i) & (ii) using these values we wont get integer values. Thus a definite solution is obtained.

Statement 2 alone is sufficient

Thanks for such a simple solution. I completely missed using the integer condition for h and k. +1 Kudos for you.

why h & k cannot be odd? Product and sum of the roots can have decimal values.

gmatclubot

Re: If h and k are both positive integers, are both roots of the equation
[#permalink]
05 Nov 2016, 05:43

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...