karthikpandian19 wrote:If h and k are both positive integers, are both roots of the equation 2x^2 + hx +k integers?

The product of the roots is 6
kh = 132

For any quadratic in the form ax² + bx + c = 0:
The sum of the roots = -b/a.
The product of the roots = c/a.

Thus, for the equation 2x²+ hx + k = 0:
The sum of the roots -h/2.
The product of the roots = k/2.
Since the sum of the roots is negative and the product of the roots is positive, both roots are negative.

Statement 1: The product of the roots is 6.
Thus, k/2 = 6, implying that k=12.
No information about h.
INSUFFICIENT.

Statement 2: kh = 132.
Substituting k = 132/h into 2x²+ hx + k = 0, we get:
2x²+ hx + 132/h = 0.

Thus:
The sum of the roots = -h/2.
The product of the roots = (132/h)/2 = 66/h.

For both roots to be integers, their sum (-h/2) and their product (66/h) will have to be integers, implying that h will have to be an even factor of 66:
h=2, h=6, h=22, or h=66.

Case 1:
If h=2, then the sum of the roots = -2/2 = -1.
It is not possible for two negative integers to have a sum of -1.

Case 2:
If h=6, then the sum of the roots = -6/2 = -3 and the product of the roots = 66/6 = 11.
No integers values will work here.

Case 3:
If h=22, then the sum of the roots = -22/2 = -11 and the product of the roots = 66/22 = 3.
No integer values will work here.

Case 4:
If h=66, then the sum of the roots = -66/2 = -33 and the product of the roots = 66/66 = 1.
No integer values will work here.

Thus, it is not possible that the roots are integers.
SUFFICIENT.

The correct answer is B.

karthikpandian19 wrote:If h and k are both positive integers, are both roots of the equation 2x^2 + hx +k integers?

The product of the roots is 6
kh = 132

Hi Karthik:

We are given 2(x^2)+hx+k=0
we can reduce this to x^2+(h/2)x+k/2 = 0
Let the roots be a and b. Then recall that if sum of roots is S, and product P,
We have x^2-Sx+P = 0 where S= (a+b) and P=ab

Here we are given, a+b=-h/2; and ab=k/2. With that in mind let's tackle this one:

1) We are given that P=ab=6, since ab=6, a=6/b, we can satisfy this for any number b to get a.
Ex: b=1, a=6, or b= 12, a=0.5, INSUFFICIENT

2) Here we are given kh=2
recall that h=-2(a+b) and k=2ab
then hk = -4(a+b)(ab) .
We are given that hk=132
then -4(a+b)(ab)=132 => (a+b)(ab) = 132/-4 = -33

We need to find whether a, b are both integers. Let a and be be integers,
then a+b is an integer, ab is an integer.

Now RHS = -33 is odd, then left hand side must be odd as well.
Product of two integers is odd if and only if each integer is odd.
So we have a+b is odd, ab is odd.

But for a+b to be odd, one of them must be even. (since odd+even=odd)
But if one of a, b is even, ab has to be even (since even*odd=even)
This is impossible given that our RHS is odd. Hence a,b can't be integers. SUFFICIENT.

Hence B is correct.

Let me know if this helps

I clearly understand the methodology, but there is one small glitch i saw, which is colored in red below:

Pls explain tht

eagleeye wrote:

karthikpandian19 wrote:If h and k are both positive integers, are both roots of the equation 2x^2 + hx +k integers?

The product of the roots is 6
kh = 132

Hi Karthik:

We are given 2(x^2)+hx+k=0
we can reduce this to x^2+(h/2)x+k/2 = 0
Let the roots be a and b. Then recall that if sum of roots is S, and product P,
We have x^2-Sx+P = 0 where S= (a+b) and P=ab

Here we are given, a+b=-h/2; and ab=k/2. With that in mind let's tackle this one:

1) We are given that P=ab=6, since ab=6, a=6/b, we can satisfy this for any number b to get a.
Ex: b=1, a=6, or b= 12, a=0.5, INSUFFICIENT

2) Here we are given kh=2
recall that h=-2(a+b) - How the negative symbol came here??? and k=2ab
then hk = -4(a+b)(ab) .
We are given that hk=132
then -4(a+b)(ab)=132 => (a+b)(ab) = 132/-4 = -33

We need to find whether a, b are both integers. Let a and be be integers,
then a+b is an integer, ab is an integer.

Now RHS = -33 is odd, then left hand side must be odd as well.
Product of two integers is odd if and only if each integer is odd.
So we have a+b is odd, ab is odd.

But for a+b to be odd, one of them must be even. (since odd+even=odd)
But if one of a, b is even, ab has to be even (since even*odd=even)
This is impossible given that our RHS is odd. Hence a,b can't be integers. SUFFICIENT.

Hence B is correct.

Let me know if this helps

karthikpandian19 wrote:I clearly understand the methodology, but there is one small glitch i saw, which is colored in red below:

Pls explain tht

eagleeye wrote:

karthikpandian19 wrote:If h and k are both positive integers, are both roots of the equation 2x^2 + hx +k integers?

The product of the roots is 6
kh = 132

Hi Karthik:

We are given 2(x^2)+hx+k=0
we can reduce this to x^2+(h/2)x+k/2 = 0
Let the roots be a and b. Then recall that if sum of roots is S, and product P,
We have x^2-Sx+P = 0 where S= (a+b) and P=ab

Here we are given, a+b=-h/2; and ab=k/2. With that in mind let's tackle this one:

1) We are given that P=ab=6, since ab=6, a=6/b, we can satisfy this for any number b to get a.
Ex: b=1, a=6, or b= 12, a=0.5, INSUFFICIENT

2) Here we are given kh=2
recall that h=-2(a+b) - How the negative symbol came here??? and k=2ab
then hk = -4(a+b)(ab) .
We are given that hk=132
then -4(a+b)(ab)=132 => (a+b)(ab) = 132/-4 = -33

We need to find whether a, b are both integers. Let a and be be integers,
then a+b is an integer, ab is an integer.

Now RHS = -33 is odd, then left hand side must be odd as well.
Product of two integers is odd if and only if each integer is odd.
So we have a+b is odd, ab is odd.

But for a+b to be odd, one of them must be even. (since odd+even=odd)
But if one of a, b is even, ab has to be even (since even*odd=even)
This is impossible given that our RHS is odd. Hence a,b can't be integers. SUFFICIENT.

Hence B is correct.

Let me know if this helps

Hi Karthik:

I didn't see any glitches. I will however copy the relevant equations here below.

For roots a and b, Sum = (a+b)=S, Product = ab = P
Standard form is (x-a)(x-b) = 0
=> x^2 - x(a+b) + ab = 0 (which can also be expressed as x^2-Sx+P=0)
Now the equation we were given was
x^2 + (h/2)*x + k/2 = 0

Comparing the x coefficients of the two equations:

-(a+b) = h/2
=> h = -2(a+b)

Now i got it from the last two sentences

thanks

eagleeye wrote:

karthikpandian19 wrote:I clearly understand the methodology, but there is one small glitch i saw, which is colored in red below:

Pls explain tht

eagleeye wrote:

karthikpandian19 wrote:If h and k are both positive integers, are both roots of the equation 2x^2 + hx +k integers?

The product of the roots is 6
kh = 132

Hi Karthik:

We are given 2(x^2)+hx+k=0
we can reduce this to x^2+(h/2)x+k/2 = 0
Let the roots be a and b. Then recall that if sum of roots is S, and product P,
We have x^2-Sx+P = 0 where S= (a+b) and P=ab

Here we are given, a+b=-h/2; and ab=k/2. With that in mind let's tackle this one:

1) We are given that P=ab=6, since ab=6, a=6/b, we can satisfy this for any number b to get a.
Ex: b=1, a=6, or b= 12, a=0.5, INSUFFICIENT

2) Here we are given kh=2
recall that h=-2(a+b) - How the negative symbol came here??? and k=2ab
then hk = -4(a+b)(ab) .
We are given that hk=132
then -4(a+b)(ab)=132 => (a+b)(ab) = 132/-4 = -33

We need to find whether a, b are both integers. Let a and be be integers,
then a+b is an integer, ab is an integer.

Now RHS = -33 is odd, then left hand side must be odd as well.
Product of two integers is odd if and only if each integer is odd.
So we have a+b is odd, ab is odd.

But for a+b to be odd, one of them must be even. (since odd+even=odd)
But if one of a, b is even, ab has to be even (since even*odd=even)
This is impossible given that our RHS is odd. Hence a,b can't be integers. SUFFICIENT.

Hence B is correct.

Let me know if this helps

Hi Karthik:

I didn't see any glitches. I will however copy the relevant equations here below.

For roots a and b, Sum = (a+b)=S, Product = ab = P
Standard form is (x-a)(x-b) = 0
=> x^2 - x(a+b) + ab = 0 (which can also be expressed as x^2-Sx+P=0)
Now the equation we were given was
x^2 + (h/2)*x + k/2 = 0

Comparing the x coefficients of the two equations:

-(a+b) = h/2
=> h = -2(a+b)