Solve the following :-

l x+1/ x l + l x + 1l = (x+1)^{2}/ l x l

we have,

But *x* = – 1 is not in the interval which we are considering in this case.

So ew do not get any solution for x< -1.

**When –1 ≤ ****x**** < 0,**

(*x* + 1)^{2} – (*x* + 1) + *x* (*x* + 1) = 0

⇒(*x* + 1) {(*x* + 1) –1 + *x*} = 0

⇒(*x* + 1) × (2*x*) = 0

⇒*x *= 0 or – 1

Since *x* ≠ 0 ⇒ *x* = – 1 is a solution when – 1 ≤ *x* < 0.

**When ****x**** ≥ 0,**

(*x* + 1)^{2} – *x* (*x* + 1) – (*x* + 1) = 0

⇒ (*x* + 1) {(*x* + 1) – *x* – 1} = 0

⇒ 0 = 0, which is true.

Thus for *x* ≥ 0 , every value of *x* will be the solution of given equation, provided that *x* ≠ 0.

Hence, the solution set of given equation is

S = {– 1} ∪ (0, ∞)

**
**