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?
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 ?