A **radical equation** or **irrational equation**, is all those that have an unknown quantity under the radical sign. To solve this type of equations we must isolate the roots that contain the unknowns in a member of the equation and then raise both members to a power equal to the root index.

**Steps to solve an equation with radicals or irrational**

To solve radical equations we must follow these steps:

- Isolate the radical containing the variable in question to one side of the equality.
- Raise both limbs to a power equal to the index of the root involved. For example, if the unknown is under a square root, we will square both sides of the equality.
- If the equation obtained in step 2 does not contain radicals, it must be solved by applying the operations (+, -, ×, ÷) that are necessary to obtain the solution value. If, on the other hand, it has one or more radicals, steps 1 and 2 are repeated until obtaining an equation for radicals.
- Substitute the values obtained in the previous step into the initial equation and evaluate if they comply with the latter. Sometimes, when doing the above, “wrong” solutions may appear, called strange roots. For this reason, the initial equation must always be checked to detect and discard the solutions that are not valid, that is, those that do not comply with it.

**Example 1**: Solve the following equation: √ (8x - 7) - 2 = 3.

We isolate to the first member the radical that contains the variable, for this we add +2 on both sides of the equality:

√ (8x - 7) - 2 + 2 = 3 +2

√ (8x - 7) = 5

Let's note that the unknown is under a square root, therefore, we square both members:

[√ (8x - 7)] ² = 5²

We remove the squared radical:

8x - 7 = 25

We add 7 to both members of the equation:

8x - 7 + 7 = 25 + 7

8x = 32

Dividing both members of the equation by 8:

8x / 8 = 32/8

x = 4

Substituting x = 4 in the original equation to evaluate whether it is a strange root or not, we check that √ (8x - 7) - 2 = 3 is correct. So:

Solution: x = 6.

**Example 2**: Solve the following equation [∛ (4x - 1) + 5] / 2 = 4

We isolate the radical containing the variable to the first member, for this we multiply both sides by 2 and then subtract 5:

{[∛ (4x - 1) + 5] / 2} × 2 = 4 × 2

∛ (4x - 1) + 5 = 8

∛ (4x - 1) + 5 - 5 = 8 - 5

∛ (4x - 1) = 3

Let's note that the unknown is under a cube root, therefore, we cube both members:

[∛ (4x - 1)] ³ = 3³

We remove the cubed radical:

4x - 1 = 27

We add 1 to both members of the equation:

4x - 1 + 1 = 27 + 1

4x = 28

Dividing both members of the equation by 4:

4x / 4 = 28/4

x = 7

Substituting x = 7 into the original equation to evaluate whether it is a strange root or not, we check that [∛ (4x - 1) + 5] / 2 = 4 is correct. So:

Solution: x = 7.

**Example 3**: Solve the following equation: √ [2 + √ (x - 5)] = √ (13 - x)

Let's note that the unknown is under a square root, therefore, we square both members:

{√ [2 + √ (x - 5)]} ² = {√ (13 - x)} ²

We remove the radicals squared:

2 + √ (x - 5) = 13 - x

We subtract 2 from both members of the equation:

√ (x - 5) = 13 - x - 2

Let's see that the unknown is still under a radical sign:

√ (x - 5) = 11 - x

Therefore, we square both members:

[√ (x - 5)] ² = (11 - x) ²

We remove the radical squared; remember that (a - b)^{2 }= to^{2 }- 2ab + b^{2 }:

x - 5 = 121 - 22x + x^{2}

We obtain a quadratic equation of quadratic:

x^{2 }- 23x + 126 = 0

When solving it, we find two possible solutions:

x = [- b ± √ (b^{2 }- 4ac)] / 2a

With a = 1, b = -23, c = 126:

x = {23 ± √ [(- 23)^{2 }- 4 (1) (126)]} / 2a

x_{1 }= [23 + √ (529 - 504)] / 2 = (23 + √25) / 2 = (23 + 5) / 2 = 28/2 = 14

x_{2 }= [23 - √ (529 - 504)] / 2 = (23 - √25) / 2 = (23 - 5) / 2 = 18/2 = 9

By substituting x_{1 }= 14 in the original equation to evaluate whether it is a strange root or not, we check that √ [2 + √ (14 - 5)] ≠ √ (13 - 14), is not true. So:

x_{1 }= 14 is a strange root

Now by substituting x_{2 }= 9 in the original equation to evaluate if it is a strange root or not, we check that √ [2 + √ (9 - 5)] = √ (9 - 14), is correct. So:

Solution: x = 31