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

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

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25 Jul 2014, 15:53
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Question Stats:

22% (03:34) correct 78% (02:14) wrong based on 587 sessions

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If h and k are both positive integers, are both roots of the equation 2x^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.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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19 Aug 2014, 21:05
8
3
prasun9 wrote:

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?

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'.

Hence this statement alone is sufficient.

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

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26 Jul 2014, 00:11
5
6
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

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

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25 Jul 2014, 23:58
1
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.

(1) Insufficient
$$(2x+3)(x+4) = \frac{3}{2}*4=6$$ One root is an integer, the other is not.

$$(2x+6)(x+2) = \frac{6}{2}*2=6$$ Both roots are Integers

(2) Sufficient
Prime Roots of $$132 = 2^2*3*11$$

Only 2 combinations are possible

If h=-3 & k=-44
$$(2x+11)(x-4)$$--> One root is an integer, the other is not.

If h=11 & k=12
$$(2x+3)(x+4)$$ --> One root is an integer, the other is not.

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Joined: 28 Apr 2014
Posts: 223
Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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26 Jul 2014, 20:47
1
justbequiet 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.

(1) Insufficient
$$(2x+3)(x+4) = \frac{3}{2}*4=6$$ One root is an integer, the other is not.

$$(2x+6)(x+2) = \frac{6}{2}*2=6$$ Both roots are Integers

(2) Sufficient
Prime Roots of $$132 = 2^2*3*11$$

Only 2 combinations are possible

If h=-3 & k=-44
$$(2x+11)(x-4)$$--> One root is an integer, the other is not.

If h=11 & k=12
$$(2x+3)(x+4)$$ --> One root is an integer, the other is not.

Why only two combinations are possible ? Why not 22,6 or 66,2
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Posts: 97
Location: Philippines
Concentration: Marketing, Entrepreneurship
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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27 Jul 2014, 00:44
himanshujovi wrote:
justbequiet 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.

(1) Insufficient
$$(2x+3)(x+4) = \frac{3}{2}*4=6$$ One root is an integer, the other is not.

$$(2x+6)(x+2) = \frac{6}{2}*2=6$$ Both roots are Integers

(2) Sufficient
Prime Roots of $$132 = 2^2*3*11$$

Only 2 combinations are possible

If h=-3 & k=-44
$$(2x+11)(x-4)$$--> One root is an integer, the other is not.

If h=11 & k=12
$$(2x+3)(x+4)$$ --> One root is an integer, the other is not.

Why only two combinations are possible ? Why not 22,6 or 66,2

Its something I overlooked when trying out the options. But regardless, result will be the same.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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29 Jul 2014, 19:55
2
1
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.
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Posts: 370
Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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18 Aug 2014, 12:48
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?
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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19 Aug 2014, 16:34
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.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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20 Aug 2014, 09:53
VeritasPrepKarishma wrote:
prasun9 wrote:

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

Please, how did you arrive at the "sum of roots" and "product of roots"? I am totally lost on this part...
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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20 Aug 2014, 10:56
1
VeritasPrepKarishma wrote:
prasun9 wrote:

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

Please, how did you arrive at the "sum of roots" and "product of roots"? I am totally lost on this part...

This is a standard formula for the roots of quadratic equations. Assume the quadratic equation is ax^2 + bx + c = 0

Sum of roots = -b/a
Product of roots = c/a
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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20 Aug 2014, 20:05
1
VeritasPrepKarishma wrote:
prasun9 wrote:

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

Please, how did you arrive at the "sum of roots" and "product of roots"? I am totally lost on this part...

Here is a post explaining how roots and coefficients are linked: http://www.veritasprep.com/blog/2012/01 ... equations/
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If h and k are both positive integers, are both roots of the equation  [#permalink]

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27 Oct 2016, 05:36
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.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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30 Oct 2016, 23:31
hubahuba wrote:
If h and k are both positive integers, are both roots of the equation 2x^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.

here is my take
2x^2 + hx + k = 0 can be expressed as x^2 + h/2 x + k/2 = 0 where , -h/2 = the sum of the roots and k/2 is the product.

from 1

k/2 = 6 , thus the roots ( x,y) product is 6 , xy = 6 = 3*2 or sqrt 3 * 2sqrt 3 ... insuff

from2

kh = 132 thus kh/4 = 3*11 , assuming roots are x,y thus xy(x+y) = 3*11 thus x+y = 3*11/xy for ( x+y) to be integer xy must either 33 , -33 , 1 or -1

but if xy = 3*11 or -3*11 then (x+y) = 1 or -1 ( impossible)

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

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31 Oct 2016, 23:21
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?

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'.

Hence this statement alone is sufficient.

VeritasPrepKarishma

Can you pls explain the highlighted part.

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

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31 Oct 2016, 23:40
gauravk wrote:
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?

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'.

Hence this statement alone is sufficient.

VeritasPrepKarishma

Can you pls explain the highlighted part.

TIA

Say h = 11
Sum of the roots = -h/2 = -11/2 = -5.5
Product of the roots = 6
The roots would be -1.5 and -4.

Both roots may not be integers.

So the first statement is not sufficient.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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05 Nov 2016, 04:43
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.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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06 Aug 2018, 04:16
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

What prevents h/2 and k/2 from having decimal values. It is not mentioned anywhere in the question or statements.
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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14 Oct 2018, 18:40
prasun9 wrote:

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?

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'.

Hence this statement alone is sufficient.

Hey, can you explain how did you arrive at Sum of roots = -h/2 Product of roots = k/2 ?
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Re: If h and k are both positive integers, are both roots of the equation  [#permalink]

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15 Oct 2018, 02:17
menonpriyanka92 wrote:
prasun9 wrote:

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?

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'.

Hence this statement alone is sufficient.

Hey, can you explain how did you arrive at Sum of roots = -h/2 Product of roots = k/2 ?

Given a quadratic equation, $$ax^2 + bx + c = 0$$
Sum of roots = -b/a
Product of roots = c/a

You can easily derive them by expanding the following:
$$ax^2 + bx + c = a(x – p)(x – q)$$ (where p and q are the roots of the equation)

$$ax^2 + bx + c = ax^2 -a(p + q)x + apq$$

$$b = - a(p + q)$$
$$(p + q) = -b/a$$

$$c = apq$$
$$pq = c/a$$
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Re: If h and k are both positive integers, are both roots of the equation &nbs [#permalink] 15 Oct 2018, 02:17
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