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805+ Level|   Algebra|      
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If h and k are both positive integers, are both roots of the equation 25 Jul 2014, 16:53
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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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If h and k are both positive integers, are both roots of the equation 19 Aug 2014, 22:05
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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.
Show more

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'.

Hence this statement alone is sufficient.

Answer (B)
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If h and k are both positive integers, are both roots of the equation 26 Jul 2014, 01:11
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hubahuba
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.
Show more




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


General Discussion
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If h and k are both positive integers, are both roots of the equation 26 Jul 2014, 00:58
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hubahuba
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.
Show more

(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.

Answer is B
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If h and k are both positive integers, are both roots of the equation 26 Jul 2014, 21:47
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justbequiet
hubahuba
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.

Answer is B
Show more

Why only two combinations are possible ? Why not 22,6 or 66,2
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If h and k are both positive integers, are both roots of the equation 27 Jul 2014, 01:44
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justbequiet
hubahuba
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.

Answer is B

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

Its something I overlooked when trying out the options. But regardless, result will be the same.
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If h and k are both positive integers, are both roots of the equation 29 Jul 2014, 20:55
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hubahuba
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


Show more

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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If h and k are both positive integers, are both roots of the equation 18 Aug 2014, 13:48
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VenoMftw



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


Show more

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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If h and k are both positive integers, are both roots of the equation 19 Aug 2014, 17:34
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hubahuba
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


Show more

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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If h and k are both positive integers, are both roots of the equation 20 Aug 2014, 10:53
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2x^2 + hx + k = 0
Sum of roots = -h/2
Product of roots = k/2

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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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If h and k are both positive integers, are both roots of the equation 20 Aug 2014, 11:56
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prasun9


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...
Show more

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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If h and k are both positive integers, are both roots of the equation Updated on: 17 Oct 2022, 01:00
Originally posted by KarishmaB on 20 Aug 2014, 21:05.
Last edited by KarishmaB on 17 Oct 2022, 01:00, edited 1 time in total.
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prasun9


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...
Show more

Check out the blog posts on the link given in my signature below.
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If h and k are both positive integers, are both roots of the equation 27 Oct 2016, 06:36
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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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If h and k are both positive integers, are both roots of the equation 31 Oct 2016, 00:31
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hubahuba
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.
Show more

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 01 Nov 2016, 00:21
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VeritasPrepKarishma


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'.

Hence this statement alone is sufficient.

Answer (B)
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VeritasPrepKarishma

Can you pls explain the highlighted part.

TIA
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If h and k are both positive integers, are both roots of the equation 01 Nov 2016, 00:40
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VeritasPrepKarishma


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'.

Hence this statement alone is sufficient.

Answer (B)

VeritasPrepKarishma

Can you pls explain the highlighted part.

TIA
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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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If h and k are both positive integers, are both roots of the equation 05 Nov 2016, 05:43
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hubahuba
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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why h & k cannot be odd? Product and sum of the roots can have decimal values.
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If h and k are both positive integers, are both roots of the equation 06 Aug 2018, 05:16
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VenoMfTw
hubahuba
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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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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menonpriyanka92
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If h and k are both positive integers, are both roots of the equation 14 Oct 2018, 19:40
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VeritasKarishma
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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'.

Hence this statement alone is sufficient.

Answer (B)
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Hey, can you explain how did you arrive at Sum of roots = -h/2 Product of roots = k/2 ?
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If h and k are both positive integers, are both roots of the equation 15 Oct 2018, 03:17
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menonpriyanka92
VeritasKarishma
prasun9


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'.

Hence this statement alone is sufficient.

Answer (B)

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