The correct option is B 1
|x2−2x−3|+4x=0
|(x+1)(x−3)|+4x=0
For xϵ(−∞,−1]∪[3,∞)
x2−2x−3+4x=0
x2+2x−3=0
(x+3)(x−1)=0
x=−3 and x=1
However x=1 does not fall in the above range.
Hence x=−3 is a possible real root.
For xϵ(−1,3)
−x2+2x+3+4x=0
−x2+6x+3=0
x2−6x−3=0
(x−3)2−12=0
(x−3)2=12
(x−3)=±2√3
x=3±2√3 ... irrational roots.
Hence there exists only one rational root that is x=−3