The correct option is D Exactly two roots of the equation are real
Given equation
2x4−3x3−x2−3x+2=0
Dividing the equation by x2,
2(x2+1x2)−3(x+1x)−1=0 ⇒2((x2+1x2)2−2)−3(x+1x)−1=0
Assuming
x+1x=t⇒t∈R−(−2,2)
Then,
2(t2−2)−3t−1=0
⇒2t2−3t−5=0⇒(t+1)(2t−5)=0⇒t=−1,52∴t=52
Now,
x+1x=52⇒2x2−5x+2=0⇒(2x−1)(x−2)=0⇒x=2,12