Let t=(3x2+4x+2)
⇒(t+1)2−(k+1)(t)(t+1)+k(t)2=0
⇒t2+2t+1−(k+1)t2−(k+1)t+kt2=0
⇒2t+1−kt−t=0
⇒t(1−k)=−1
Replacing the value of t again
3x2+4x+2=t=k−11
⇒3x2+4x+(k−12k−3)=0
For real roots, D≥0
⇒16−4×3×(k−12k−3)≥0
⇒k−16k−9−4≤0
⇒(k−1)(2k−5)≤0
⇒k∈(1,25]