The equationx+1−x−1=4x−1,(x∈R)

The given equation is valid if x + 1 ≥ 0, x – 1≥ 0 and 4x – 1 ≥ 0 i.e. if x ≥ 1. Squaring both the sides we get                   x+1+x−1−2(x+1)(x−1)=4x−1⇒                  1−2x=2(x+1)(x−1)Squaring again, we get                      1−4x+4x2=4x2−1⇒                 4x=5 or x=5/4.Putting this value of x in the given equation, we get                      54+1−54−1=454−1⇒                 32−12=2 or 1=2which is not true.Thus, the given equation has no solution.