If x is real then x2−2x+4x2+2x+4 takes values in the interval
[1/3,3]
(1/3,3)
(3,3)
(−1/3,3)
Let y=x2−2x+4x2+2x+4.Then,
x2(y−1)+2x(y+1)+4(y−1)=0
⇒ 4(y+1)2−16(y−1)2≥0 [∵x is real ]
⇒ (y+1)2−{2(y−1)}2≥0⇒ (3−y)(3y−1)≥0⇒1/3≤y≤3