# Solve equation (x^2-2x+1)^(1/2)=x+1?

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### 2 Answers

You need to replace `(x - 1)^2` for expansion `x^2 - 2x + 1` , such that:

`sqrt ((x - 1)^2) = x + 1`

Taking the square root to the left side, yields:

`+-(x - 1) = x + 1`

You need to solve for `x` the equation, considering the left side positive, such that:

`x - 1 = x + 1 => - 1 = 1` invalid

You need to solve for x the equation, considering the left side negative, such that:

`-x + 1 = x + 1`

You need to isolate the terms that contain `x` to the left side, such that:

`-x - x = -1 + 1 => -2x = 0 => x = 0`

You need to test the value `x = 0` in equation, such that:

`sqrt(0^2 - 2*0 + 1) = 0 + 1 => sqrt 1 = 1 => 1 = 1` valid

**Hence, evaluating the solution to the given equation, yields **`x = 0.`

The equation (x^2-2x+1)^(1/2)=x+1 has to be solved for x.

(x^2-2x+1)^(1/2)=x+1

take the square of both the sides.

((x^2-2x+1)^(1/2))^2=(x+1)^2

(x^2-2x+1)=(x+1)^2

x^2 - 2x + 1 = x^2 + 2x + 1

-2x + 1 = 2x + 1

4x = 1 - 1

4x = 0

x = 0

The solution of the given equation is x = 0