Irrational equations are equations in which the unknown \(x\) appears under the sign of a root or is raised to power with a fractional exponent. They represent a class of equations that can be solved through specific methods, considering the properties of roots and exponents.

\[\sqrt[\Large{n}]{f(x)} = g(x) \quad \text{or} \quad f(x)^{\Large{\frac{1}{n}}} = g(x) \]

where \(f(x)\) and \(g(x)\) are polynomials of any degree with real coefficients.

The key distinction between rational and irrational equations lies in their structure. Rational equations consist of fractions with polynomials in both numerator and denominator and can always be expressed as a ratio of two polynomials. Irrational equations feature roots of various orders and solutions that cannot be described as rational numbers. These roots are represented using radical notation and an index that indicates their order. This fundamental difference is crucial to understanding the nature of these equations. For example:


is an irrational equation since the variable is inside a root.


is a rational equation since the variable \(x\) appears in a polynomial of the first degree in the denominator.

Irrational equations come in various forms, which is useful to recognize in order to solve them easily. Let’s see below what they are.

Irrational equations with even index

Irrational equations in which the index \(n\) of the root is even have the form:

\[\sqrt[\large{n}]{f(x)} = g(x) \quad n \in 2\mathbb{Z}\]

where \(2\mathbb{Z}\) is the set of even integers. In this case, one must impose the following conditions of existence.

  • When the equation is of the form: \[\sqrt[\large{n}]{f(x)} = k\] where \(k\) is a positive or null real number, the field of existence of the unknown \(x\) is given by the values of the variable such that \(f(x) \geq 0\).

  • When the equation is of the form: \[\sqrt[\large{n}]{f(x)} = k\] where \(k\) is a negative real number, the equation has no solution since a root of even index always results in a non-negative number.

  • When the equation is of the form: \[\sqrt[\large{n}]{f(x)} = g(x)\] where \(g(x)\) is a rational function, the field of existence of the unknown \(x\) is given by the resolution of the system of inequalities: \[ \begin{cases} f(x) \geq 0 \\[2ex] g(x) \geq 0 \\[2ex] f(x)=g(x)^n \end{cases}\]

Irrational equations with odd index

If the root’s index \(n\) is an odd number, there is no need to impose any existence conditions, as the root’s field of existence is the set of real numbers. In other words, in this case, it is always possible to extract the roots of negative numbers.

\[\sqrt[\Large{3}]{-8} = -2\]

Solving Method

To solve an irrational equation, follow these steps:

  • Determine the field of existence by considering the even or odd index of its component roots, and impose the relevant conditions seen above.

  • Eliminate the roots by raising the members of the equation to a power.

  • Perform the calculations and determine the solutions. Check their admissibility and verify if they satisfy the initial equation.

It’s essential to remember that when solving equations with roots, the roots on either side of the equal sign should be kept separate to avoid generating further roots due to power elevation. For instance, consider the following irrational equation:

\[\sqrt{2x} -\sqrt{x +1} = 0\]

  • When we square a root, it creates a new root that can make the calculation more complicated: \[ \begin{align} & (\sqrt{2x} -\sqrt{x +1} )^2 = 0 \\[2ex] & 2x -(x +1) + 2\sqrt{2x}\cdot\sqrt{x+1} = 0\\[2ex] \end{align} \]

  • In order to simplify calculations, it is preferable to separate the roots on either side of the equals sign where possible: \[ \begin{align} & \sqrt{2x} = \sqrt{x +1} \\[2ex] & (\sqrt{2x})^2 = (\sqrt{x +1})^2 \\[2ex] & 2x = x +1 \\[2ex] & 2x-x = 1 \\[2ex] & x = 1 \end{align} \]

Solving these equations is not difficult, but lack of practice may make them seem more complex than they actually are once you understand their behavior.


Solve the irrational equation:

\[\sqrt{x^2-2} = \sqrt{4x} \]

The equation is of the form: \[\sqrt[\large{n}]{f(x)} = g(x)\] where \(n\), the index of the root is even. The admissible set of solutions is determined by solving the system of inequalities:\[ \begin{cases} f(x) \geq 0 \\[2ex] g(x) \geq 0 \\ \end{cases}\]

Substitute (f(x)) and (g(x)) for the polynomials under the roots to obtain: \[ \begin{cases} x^2-2 \geq 0 \\[2ex] 4x \geq 0 \\ \end{cases} \]

Solve the second-degree equation associated with the first inequality \(x^2-2 \geq 0\) and find its solutions. In this case, we have: \[x_{1,2} = \pm \sqrt{2} \] therefore, the field of existence of the inequality is given by the intervals: \[(-\infty, -\sqrt{2}] \ \cup \ [\sqrt{2}, +\infty)\]

For the second inequality we have: \[x \geq 0\]

The intersection of the intervals found gives the admissible set of solutions. We can use the graphical method to determine it visually:

\[ -\sqrt{2}\]\[ 0\]\[ +\sqrt{2}\]

The system admits solutions in the interval \([\sqrt{2}, +\infty)\).

We will now solve the original equation by isolating the roots to the left and right of the equal sign, and then squaring both members. We have: \[ \begin{align} & \sqrt{x^2 -2} = \sqrt{4x} \\[0.5em] & x^2 -2 = 4x\\[0.5em] & x^2 -4x -2 = 0 \end{align} \]

We have obtained a second-degree equation that we can solve using the quadratic formula. \[ \begin{align} x_{1,2} &= \frac{-(-4) \pm \sqrt{(-4)^2 -4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \\[1ex] & = \frac{4 \pm \sqrt{16 +8}}{2} \\[1ex] & = \frac{4 \pm \sqrt{24}}{2} \\[1ex] & = \frac{4 \pm 2\sqrt{6}}{2} \\[1ex] & = 2 \pm \sqrt{6} \\[1ex] \end{align} \]

Both solutions found are acceptable since they fall within the admissible set of solutions for the initial equation. Finally, we need to verify if they satisfy the equation.

For \(x = 2 +\sqrt{6}\) we have: \[ \begin{align} & \sqrt{(2 +\sqrt{6})^2 -2} = \sqrt{4\cdot(2+\sqrt{6})} \\[0.5em] & \sqrt{4 +4\sqrt{6} +6 -2}= \sqrt{8 +4\sqrt{6}} \\[0.5em] & \sqrt{8 +4\sqrt{6}} = \sqrt{8 +4\sqrt{6}}\\[0.5em] \end{align} \]

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

For \(x = 2 -\sqrt{6}\) we have: \[ \begin{align} & \sqrt{(2 -\sqrt{6})^2 -2} = \sqrt{4\cdot(2 -\sqrt{6})} \\[0.5em] & \sqrt{4 -4\sqrt{6} +6 -2}= \sqrt{8 -4\sqrt{6}} \\[0.5em] & \sqrt{8 -4\sqrt{6}} = \sqrt{8 -4\sqrt{6}}\\[0.5em] \end{align} \]

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

The solution to the equation is:

\[x_1= 2 -\sqrt{6} \quad \quad x_2= 2 +\sqrt{6} \]