The correct option is B x2−12x+35=0
Given: (1−m)x2−(2−3m)x−3m=0
On comparing with standard form of quadratic equation ax2+bx+c=0
we get, a=(1−m),b=−(2−3m),c=−3m
If roots are distinct integers, then discriminant should be perfect square of an integer.
D=b2−4ac=(−(2−3m))2−4⋅(1−m)⋅(−3m)
⇒D=4+9m2−12m+12m−12m2
⇒D=−3m2+4
Also, D>0⇒−3m2+4>0
⇒3m2−4<0
⇒(√3m−2)(√3m+2)<0
⇒m∈(−2√3,2√3)
Since, m∈W⇒m∈{0,1}
If m=0,D=−3(0)2+4=4, which is a perfect square.
Similarly, if m=1,D=−3(1)2+4=1 which is a perfect square.
Hence, the possible values of m are 0,1.
When m=0, the quadratic equation becomes x2−2x=0
⇒α=0;β=2 ⋯(i)
When m=1, the equation becomes x−3=0 which is not a quadratic equation since coefficient of x2 is 0
Thus, we have α=0;β=2
Now, we need to find the equation whose roots are α+5 & β+5
∴roots are 0+5=5,2+5=7
∴The equation is (x−5)(x−7)=0
⇒x2−12x+35=0