# Irrational Equation A2
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/irrational-equation-a-2/
Fetched from algebrica.org test 7192; source modified 2025-04-25T18:11:10.
Before you begin, review the key concepts behind [irrational equations](../irrational-equations.md) to better follow the solution process.
Solve the irrational equation:
\\[
\\sqrt{2x-x^2} = x-2
\\]
* * *
The equation is of the form:
\\[
\\sqrt\[n\]{f(x)} = g(x)
\\]
where \\(n\\), the index of the root is even. The domain (or set of admissible solutions) is determined by solving the [system of inequalities](../systems-of-inequalities.md):
\\[
\\begin{cases} f(x) \\geq 0 \\\\\[2ex\] g(x) \\geq 0 \\end{cases}
\\]
The system becomes:
\\[
\\begin{cases} 2x-x^2 \\geq 0 \\\\\[2ex\] x-2 \\geq 0 \\\\ \\end{cases}
\\]
* * *
For \\(x-2 \\geq 0\\) we have \\(x \\geq 2\\).
For \\( 2x-x^2 \\geq 0\\), since the sign of the second-degree term is negative, multiply everything by -1 and invert the sign of the inequality. We obtain:
\\[
x^2-2x \\leq 0
\\]
* * *
Solve the [second-degree](../quadratic-equations.md) equation associated with the inequality \\(x^2-2x \\leq 0\\) and find its solutions. The equation becomes:
\\[
\\begin{align} & x^2-2x = 0 \\\\\[1ex\] & x(x-2) = 0 \\end{align}
\\]
The solution to the equation is \\(x=0\\) and \\(x=2\\). Since the sign of the [inequality](../inequalities-with-absolute-value.md) is less than 0, the set of solutions that satisfy the inequality is between 0 and 2.
The admissible set of solutions is given by the intersection of the intervals found for \\(f(x)\\) and \\(g(x)\\). Use the graphical method to determine it visually:
\\[
0
\\]
\\[
2
\\]
The only admissible solution of the initial equation is \\(x = 2\\). Being the only possible value, we could stop here and verify if it satisfies the initial equation. The following steps are shown for completeness.
* * *
Solve the initial equation by squaring both members. We have:
\\[
\\begin{align} &2x -x^2 = (x-2)^2 \\\\\[1ex\] &2x -x^2 = x^2 -4x +4 \\\\\[1ex\] &2x -x^2 - x^2 +4x -4 = 0\\\\\[1ex\] &-2x^2 +6x -4 = 0\\\\\[1ex\] &x^2 -3x -2 = 0 \\end{align}
\\]
* * *
The second-degree equation can be [factorized](../factoring-quadratic-equations.md) as:
\\[
(x-1)(x-2) = 0
\\]
The solution to the equation is \\(x= 1\\) and \\(x = 2\\).
* * *
The only acceptable solution is \\( x = 2 \\), since it falls within the domain of the original equation. Now verify whether it satisfies the original equation. This step is crucial, because a value may belong to the domain but still not satisfy the equation, in which case it must be discarded as an extraneous solution.
For \\(x = 2\\) we have:
\\[
\\begin{align} &\\sqrt{2\\cdot 2 -(2)^2} = 2-2 \\\\\[1ex\] &\\sqrt{4 -4} = 0 \\\\\[1ex\] &0 = 0 \\end{align}
\\]
The equality is verified, so \\(x\\) is a solution to the equation.
The solution to the equation is:
\\[
x= 2
\\]
