The correct option is D 4
Given equation is x2−13x+6c=0
On comparing with the standard quadratic equation ax2+bx+c=0,a≠0
we get, a=1; b=−13; c=6c
Using discriminant method,
x=−b±√D2a
If roots are distinct integers, then discriminant should be perfect square of an integer.
Now, D=b2−4ac=(−13)2−4.1.6c
D=169−24c
Also, D>0⇒169−24c>0⇒c<16924
Since, c∈N⇒c must be an element of set {1,2,3,4,5,6,7}
So, If c=1, D=169−24(1)=145, which is not a perfect square.
c=2, D=169−24(2)=121, it is a perfect square .
c=3, D=169−24(3)=97, which is not a perfect square.
c=4, D=169−24(4)=73,which is not a perfect square.
c=5, D=169−24(5)=49 it is a perfect square .
c=6, D=169−24(6)=25, it is a perfect square .
c=7, D=169−24(7)=1, it is a perfect square .
Hence, the possible values are 2,5,6,7. ∴ The number of possible values of c are 4.