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

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If h and k are both positive integers, are both roots of the [#permalink]  25 Jul 2014, 15:53
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18% (02:47) correct 82% (01:55) wrong based on 135 sessions
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.
[Reveal] Spoiler: OA
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Kudos [?]: 25 [0], given: 504

Re: If h and k are both positive integers, are both roots of the [#permalink]  25 Jul 2014, 23:58
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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Kudos [?]: 7 [1] , given: 41

Re: If h and k are both positive integers, are both roots of the [#permalink]  26 Jul 2014, 00:11
1
KUDOS
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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Kudos [?]: 26 [1] , given: 46

Re: If h and k are both positive integers, are both roots of the [#permalink]  26 Jul 2014, 20:47
1
KUDOS
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
Manager
Joined: 04 Sep 2012
Posts: 77
Location: Philippines
Concentration: Marketing, Entrepreneurship
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GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE: Sales (Manufacturing)
Followers: 1

Kudos [?]: 25 [0], given: 504

Re: If h and k are both positive integers, are both roots of the [#permalink]  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.
Intern
Joined: 06 Feb 2014
Posts: 26
Location: United States
Concentration: Strategy, Operations
GRE 1: 1380 Q780 V600
GPA: 4
WE: Management Consulting (Aerospace and Defense)
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Kudos [?]: 22 [1] , given: 54

Re: If h and k are both positive integers, are both roots of the [#permalink]  29 Jul 2014, 19:55
1
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.
Intern
Joined: 14 Mar 2014
Posts: 8
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Kudos [?]: 7 [0], given: 41

Re: If h and k are both positive integers, are both roots of the [#permalink]  30 Jul 2014, 03:41
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:)

Senior Manager
Joined: 07 Apr 2012
Posts: 464
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Kudos [?]: 15 [0], given: 58

Re: If h and k are both positive integers, are both roots of the [#permalink]  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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Kudos [?]: 46 [0], given: 18

Re: If h and k are both positive integers, are both roots of the [#permalink]  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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Kudos [?]: 6792 [2] , given: 177

Re: If h and k are both positive integers, are both roots of the [#permalink]  19 Aug 2014, 21:05
2
KUDOS
Expert's post
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 [#permalink]  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...
Manager
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Kudos [?]: 46 [1] , given: 18

Re: If h and k are both positive integers, are both roots of the [#permalink]  20 Aug 2014, 10:56
1
KUDOS
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 [#permalink]  20 Aug 2014, 20:05
1
KUDOS
Expert's post
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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Re: If h and k are both positive integers, are both roots of the   [#permalink] 20 Aug 2014, 20:05
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