2x3−9x9+12x−k−0→(1)
To have a double root the equation would be 2(x−a)2(x−b) where a is the double root.
i.e.
2(x2−2ax+a2)(x−b)=2x3−4ax2+2a2x−2x2+4abx−2a2b
=2x3−2(2a+b)x2+2(a2+2ab)x+2(a2b)→(2)
Compare eqs.(1) and (2)
2(2a+b)=9→(3),2a(a+2b)=12→(4),k=2a2(b)
Solving (3) for b
b=9/2−2a
substituting into (4)
2a(a+2(9/2−2a))=12
2a(a+9−4a)=12
a(−3a+9)=6
a(a−3)=−2
a2−3a+2=0
(a−2)(a−1)=0
so a=2 or a=1
b=9/2−2(2)=1/2 or b=9/2−2(1)=5/2
For a=2,b=1/2
Eq. (3) 2(2(2)+(1/2))=2(4+1/2)=2(9/2)=9
Eq. (4) 2(2)(2+2(1/2))=4(3)=12
For a=1,b=5/2
Eq. (3) 2(2(1)+(5/2))=2(2+5/2)=2(9/2)=9
Eq. (4) 2(1)(1+2(5/2))=−2(6)=12
So k=2a2(b)=2(2)2(1/2)=4 or k=2(12)(5/2)=5