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 to better follow the solution process.
Solve the irrational equation:
The equation is of the form:
where $n$, the index of the root is even. The domain (or set of admissible solutions) is determined by solving the system of inequalities:
The system becomes:
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:
Solve the second-degree equation associated with the inequality $x^2-2x \leq 0$ and find its solutions. The equation becomes:
The solution to the equation is $x=0$ and $x=2$. Since the sign of the inequality 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:
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:
The second-degree equation can be factorized as:
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:
The equality is verified, so $x$ is a solution to the equation.
The solution to the equation is: