# Irrational Equation A1

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/irrational-equation-a-1/
Fetched from algebrica.org test 7188; source modified 2025-04-18T21:02:13.

Before you begin, review the key concepts behind [irrational equations](../irrational-equations.md) to better follow the solution process.

Solve the irrational equation:

\\[
\\sqrt{x^2 - 2x+1} = \\sqrt{3}
\\]

* * *

As a first step, we must determine the conditions of existence by imposing:

\\[
x^2 - 2x+1 \\geq 0
\\]

Solve the [second-degree](../quadratic-equations.md) equation associated with the inequality \\(x^2 - 2x+1 \\geq 0\\) and find its solutions. In this case, we have a [notable product](../notable-products.md) and the equation becomes:

\\[
(x - 1)(x - 1) = (x-1)^2
\\]

 The square of any real number is always non-negative value. Therefore, \\((x - 1)^2 ≥ 0\\) is true for all real values of x.

\\[
(-\\infty, +\\infty)
\\]

* * *

Proceed to solve the original equation. First, isolate the roots to the left and right of the equal sign, and then square both members. We have:

\\[
\\begin{align} & \\sqrt{x^2 - 2x+1} = \\sqrt{3} \\[0.5em] & x^2 - 2x + 1 = 3\\[0.5em] & x^2 - 2x + 1 - 3 = 0\\[0.5em] & x^2 - 2x - 2 = 0\\[0.5em] \\end{align}
\\]

* * *

We have obtained a [second-degree equation](../quadratic-equations.md) that we can solve using the [quadratic formula](../quadratic-formula.md).

\\[
\\begin{align} x\_{1,2} &= \\frac{-(-4) \\pm \\sqrt{(-2)^2 -4 \\cdot 1 \\cdot (-2)}}{2 \\cdot 1} \\\\\[1ex\] & = \\frac{4 \\pm \\sqrt{4 +8}}{2} \\\\\[1ex\] & = \\frac{4 \\pm \\sqrt{12}}{2} \\\\\[1ex\] & = \\frac{4 \\pm 2 \\sqrt{3}}{2} \\\\\[1ex\] \\end{align}
\\]

We have:

\\[
x= 1 +\\sqrt{3}
\\]



\\[
x= 1 -\\sqrt{3}
\\]

Both solutions are acceptable since they fall within the field of existence of the initial equation.

* * *

Finally, we need to verify whether they satisfy the original equation. This step is crucial because some solutions may belong to the set of possible solutions but fail to satisfy the given equation.

For \\(x = 1 +\\sqrt{3} \\) we have:

\\[
\\begin{align} &\\sqrt{(1 +\\sqrt{3})^2 -2(1 +\\sqrt{3}) +1} = \\sqrt{3}\\[0.5em] &\\sqrt{1 +2\\sqrt{3} +3 -2 -\\sqrt{3} +1} = \\sqrt{3}\\[0.5em] &\\sqrt{3} = \\sqrt{3}\\\\ \\end{align}
\\]

 The equality is verified, so \\(x\\) is a solution to the equation.

* * *

For \\(x = 1 -\\sqrt{3} \\) we have:

\\[
\\begin{align} &\\sqrt{(1 -\\sqrt{3})^2 -2(1 -\\sqrt{3}) +1} = \\sqrt{3}\\[0.5em] &\\sqrt{1 -2\\sqrt{3} +3 -2 + \\sqrt{3} +1} = \\sqrt{3}\\[0.5em] &\\sqrt{3} = \\sqrt{3}\\\\ \\end{align}
\\]

 The equality is verified, so \\(x\\) is a solution to the equation.

The solution to the equation is:

\\[
x= 1 +\\sqrt{3} \\quad \\quad x= 1 -\\sqrt{3}
\\]