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by vikram4689 » Wed Jun 20, 2012 11:24 pm
If h and k are both positive integers, are both roots of the equation 2*(x^2)+hx+k=0 integers?

a) The product of the roots is 6 .
b) kh=132

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.
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by Anurag@Gurome » Wed Jun 20, 2012 11:28 pm
Where is the equation?
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by vikram4689 » Thu Jun 21, 2012 4:15 am
added :)
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by Anurag@Gurome » Thu Jun 21, 2012 4:45 am
vikram4689 wrote:If h and k are both positive integers, are both roots of the equation 2*(x^2)+hx+k=0 integers?

a) The product of the roots is 6 .
b) kh=132
2x² + hx + k = 0 --> x² + (h/2)x + k/2 = 0 ........... (A)

Say, the roots of the equation are a and b.
Hence, (x - a)(x - b) = 0 ---> (x² -ax - bx + ab) = 0 ........(B)

Comparing the coefficients of (A) and (B),
  • h = -2(a + b)
    k = 2ab
Statement 1: k = 2*6 = 12
Only from this information we cannot conclude whether a and b are integers or not.

Not sufficient

Statement 2: kh = 132
Hence, [2ab]*[-2(a + b)] = 132 --> ab(a + b) = -33
If a and b are integers, then both ab and (a + b) will be integers.
As 33 is odd, both (a + b) and ab must be odd. But that is not possible.
Hence, a and b are not integers.

Sufficient

The correct answer is B.
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by eagleeye » Thu Jun 21, 2012 5:30 am
Hi vikram:

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.

Let me know if this helps :)
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by GMATGuruNY » Thu Jun 21, 2012 7:05 am
vikram4689 wrote:If h and k are both positive integers, are both roots of the equation 2*(x^2)+hx+k=0 integers?

a) The product of the roots is 6.
b) 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.
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by dhonu121 » Fri Jun 22, 2012 6:34 am
Statement 2: kh = 132
Hence, [2ab]*[-2(a + b)] = 132 --> ab(a + b) = -33
If a and b are integers, then both ab and (a + b) will be integers.
As 33 is odd, both (a + b) and ab must be odd. But that is not possible.
Hence, a and b are not integers.
You used the contradiction method here to prove insufficiency.
I was able to reach upto this point but wasnt able to deduce further the values of a and b.

I have a curious question:
If say a+b = Integer and/or ab=Integer, Can we deduce that a and b are both integers ?

Thanks in Advance.
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by veneetvishal » Fri Jun 22, 2012 7:07 am
dhonu121 wrote:
Statement 2: kh = 132
Hence, [2ab]*[-2(a + b)] = 132 --> ab(a + b) = -33
If a and b are integers, then both ab and (a + b) will be integers.
As 33 is odd, both (a + b) and ab must be odd. But that is not possible.
Hence, a and b are not integers.
You used the contradiction method here to prove insufficiency.
I was able to reach upto this point but wasnt able to deduce further the values of a and b.

I have a curious question:
If say a+b = Integer and/or ab=Integer, Can we deduce that a and b are both integers ?

Thanks in Advance.
Yes it is true.
We can deduce a & b are integers when a+b and ab are integer values.
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by Anurag@Gurome » Fri Jun 22, 2012 7:08 am
dhonu121 wrote:If say a+b = Integer and/or ab=Integer, Can we deduce that a and b are both integers ?
No, we can't.
Here is the proof.

Say, (a + b) = m and ab = n, where m and n are both integers.
Exploiting the theories of quadratic equations as we have done earlier in this question, we can say a and b are roots of the equation (x² - mx + n) = 0

Now, roots of this equation will be given as [m ± √(m² - 4n)]/2
We can see that even m and n are integers, a and b may not be integer.

To show with an example, say m = 3 and n = 1, then
  • a = [m + √(m² - 4n)]/2 = [3 + √(3² - 4)]/2 = (3 + √5)/2 --> Not integer
    b = [m - √(m² - 4n)]/2 = [3 - √(3² - 4)]/2 = (3 - √5)/2 --> Not integer

    Cross check
    --> (a + b) = (3 + √5 + 3 - √5)/2 = 6/2 = 3
    --> ab = (3 + √5)*(3 - √5)/4 = (9 + 3√5 - 3√5 - 5)/4 = 1
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by dhonu121 » Fri Jun 22, 2012 12:20 pm
Got it!..Thanks.
That was a wonderful explanation.
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