Irrational Equations

An irrational equation is an equation in which the unknown variable appears under a radical sign. $$ \sqrt[n]{a(x)}=b(x) $$ Here, n is an integer greater than or equal to two (n≥2).

How to solve an irrational equation

The solution method depends on the index n of the radical.

If n is odd, the equation can be solved directly by raising both sides to the n-th power: $$ [\sqrt[n]{a(x)}]^n=[b(x)]^n $$
If n is even, the equation must be solved through an equivalent system of inequalities: $$ \begin{cases} a(x) \ge 0 \\ b(x) \ge 0 \\ a(x) = [b(x)]^2 \end{cases}$$
- Verification by substitution
  Substitute each proposed root into the original equation and verify whether both sides yield the same value.
- Verification by domain conditions
  Impose the necessary conditions that ensure the equation ⁿ√a(x)=b(x) is defined. For example, the radicand a(x)≥0 must be nonnegative, and the right-hand side must also be nonnegative, b(x)≥0.

A practical example
General proof

A practical example

Consider the following irrational equation:

$$ \sqrt{x^2-3x+2}=2-x $$

The radical has an even index (n=2).

To determine the solution, we must solve the associated system of inequalities:

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ 2-x \ge 0 \\ \\ x^2-3x+2=(2-x)^2 \end{cases} $$

Let’s simplify each expression algebraically:

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ -x \ge -2 \\ \\ x^2-3x+2=4-4x+x^2 \end{cases} $$

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ x \le 2 \\ \\ x=2 \end{cases} $$

The inequality x²-3x+2≥0 holds for (-∞,1]∪[2,+∞), that is, for x≤1 or x≥2.

$$ \begin{cases} x \le 1 \ ∨ \ x \ge 2 \\ \\ x \le 2 \\ \\ x=2 \end{cases} $$

Note. The inequality x²-3x+2≥0 corresponds to the quadratic equation x²-3x+2=0, which represents an upward-opening parabola with two real roots: $$ x = \frac{-(-3) \pm \sqrt{9-8} }{2} = \frac{3 \pm 1}{2} = \begin{cases} x_1 = 1 \\ \\ x_2 = 2 \end{cases} $$

The solution set of the system contains a single element:

$$ S = \{ \ 2 \ \} $$

Therefore, x=2 is the valid solution of both the system and the original irrational equation.

Alternative solving method

Since this equation involves an even-index radical, we can also apply an equivalent alternative procedure.

$$ \sqrt{x^2-3x+2}=2-x $$

Raise both sides to the second power:

$$ [ \sqrt{x^2-3x+2}]^2=(2-x)^2 $$

$$ x^2-3x+2=(2-x)^2 $$

$$ x^2-3x+2=4-4x+x^2 $$

This step removes the radical sign.

Next, find all potential roots:

$$ a(x)=b(x) \ ∨ \ a(x) = - b(x) $$

where a(x)=x²-3x+2 and b(x)=4-4x+x².

a(x)=b(x)
$$ x^2-3x+2 = 4-4x+x^2 $$ $$ x^2-x^2-3x+4x = 4-2 $$ $$ x = 2 $$
Note. One admissible root of the equation is x=2.
a(x)=-b(x)
$$ x^2-3x+2 = - (4-4x+x^2) $$ $$ x^2-3x+2 = -4+4x-x^2 $$ $$ x^2+x^2-3x-4x +4+2 = 0 $$ $$ 2x^2-7x + 6 = 0 $$ $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ $$ x= \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(6)}}{2(2)} $$ $$ x= \frac{7 \pm \sqrt{49-48}}{4} $$ $$ x= \frac{7 \pm 1}{4} $$ $$ x = \begin{cases} \frac{7-1}{4} = \frac{3}{2} \\ \\ \frac{7+1}{4} = 2 \end{cases} $$
Note. This yields two possible roots, x=2 and x=3/2, one of which (x=2) is already known.

The possible roots of the equation are therefore:

$$ S = \{ \ 2 \ , \ \frac{3}{2} \ \} $$

We now determine which of these are admissible solutions.

Two validation approaches are available: direct substitution or checking the domain conditions.

A] Verification by substitution

Substitute each candidate value x={2,3/2} into the original equation:

$$ \sqrt{x^2-3x+2}=2-x $$

If both sides yield identical results, the root is valid; otherwise, it must be discarded.

For x=2: $$ \sqrt{2^2-3 \cdot 2+2}=2-2 $$ $$ \sqrt{4-6+2}=0 $$ $$ \sqrt{0}=0 $$ $$ 0 = 0 $$ Both sides coincide, so x=2 is valid.
For x=3/2: $$ \sqrt{(\frac{3}{2})^2-3 \cdot \frac{3}{2}+2}=2-\frac{3}{2} $$ $$ \sqrt{\frac{9}{4}-\frac{9}{2}+2}=\frac{4-3}{2} $$ $$ \sqrt{\frac{9-18+8}{4}}=\frac{1}{2} $$ $$ \sqrt{\frac{-1}{4}}=\frac{1}{2} $$ The result under the radical is negative, so this root is not admissible in the real domain.

Therefore, only x=2 is an admissible solution.

$$ S = \{ \ 2 \ \} $$

B] Verification by domain conditions

Now determine the conditions ensuring that both sides of the equation are defined in ℝ:

$$ \sqrt{x^2-3x+2}=2-x $$

The radicand x²-3x+2 ≥ 0 must be nonnegative, and the right-hand side 2-x≥0 must also be nonnegative.

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ 2-x \ge 0 \end{cases} $$

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ -x \ge -2 \end{cases} $$

$$ \begin{cases} x^2-3x+2 \ge 0 \\ \\ x \le 2 \end{cases} $$

The associated quadratic equation x²-3x+2=0 represents a parabola opening upwards.

The roots are:

$$ x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(2)} }{2} = \frac{3 \pm \sqrt{9-8} }{2} = \frac{3 \pm 1 }{2} = \begin{cases} x=1 \\ \\ x= 2 \end{cases} $$

Hence, the inequality x²-3x+2≥0 is satisfied for x≤1 or x≥2.

$$ \begin{cases} x \le 1 \ ∨ \ x \ge 2 \\ \\ x \le 2 \end{cases} $$

Now verify which of the roots S={2,3/2} satisfy these conditions.

The root x=2 satisfies all domain conditions and is therefore admissible.
The root x=3/2 does not satisfy the conditions and must be excluded.

In conclusion, the only valid root is x=2.

$$ S = \{ \ 2 \ \} $$

General proof

Consider a general irrational equation:

$$ \sqrt[n]{a(x)}=b(x) $$

By definition, the index n of the radical is an integer greater than or equal to two:

$$ n \ge 2 $$

We now distinguish between two cases: when n is odd and when n is even.

A] The index n is odd

If n is odd, the equation can be solved directly.

It is sufficient to raise both sides to the n-th power to remove the variable from under the radical:

$$ [\sqrt[n]{a(x)}]^n=[b(x)]^n $$

$$ a(x)=b(x)^n $$

Example. In this equation, the variable appears inside a cube root (n=3): $$ \sqrt[3]{x^3+x+1} = x $$ To eliminate the radical, raise both sides to the third power: $$ [ \sqrt[3]{x^3+x+1} ]^3 = [ x ]^3 $$ Since the index is odd, the cube root yields a unique real root. $$ x^3+x+1 = x^3 $$ $$ x = -1 $$

B] The index n is even

If n is even, the analysis becomes more delicate.

A radical with an even index cannot have a negative radicand in the set of real numbers.

Note. Raising any negative real number to an even power yields a positive result: $$ (-3)^2 = (-3) \cdot (-3) = 9 $$ Therefore, the square root of a negative number does not exist in ℝ. $$ \sqrt[2]{-3} = \ \text{undefined} $$ To evaluate it, one would need to introduce complex numbers, which lies outside the real domain.

Thus, the condition of existence for the radical is a(x)≥0:

$$ \begin{cases} \sqrt[n]{a(x)}=b(x) \\ \\ a(x) \ge 0 \end{cases} $$

Since b(x) is raised to an even power, it must also be nonnegative: b(x)≥0.

We therefore include this additional condition in the system:

$$ \begin{cases} \sqrt[n]{a(x)}=b(x) \\ \\ a(x) \ge 0 \\ \\ b(x) \ge 0 \end{cases} $$

This system is equivalent to the original irrational equation and defines its domain of validity.