Radical equations

irrational equationA radical equation or irrational equation, is any one that has an unknown quantity under the radical sign (Figure I). To solve such equations, the roots containing the unknowns must be isolated in one member of the equation and then raise both members to a power equal to the root index. Sometimes, when doing the above, “wrong” solutions may appear, called strange roots. Therefore, always check the initial equation to detect and discard invalid solutions.

Solving equations with radicals

The equation »√x = a» is said to be a radical equation or one irrational equation. In these, the unknowns they are under a radical sign, that is, in the part subradical. To solve an irrational equation it is necessary remove the radical sign, which is achieved by raising the two members of the equation to a power equal to the root index:

Steps to solve an equation with radicals or irrational

To solve this type of equations the following steps are carried out:

  1. Group the terms of the variable that are under the radical sign in one member of the equation and the rest of the terms in the other.
  2. Raise both limbs to a power equal to the index of the root involved.
  3. If the equation obtained in step 2 does not contain radicals, it must be solved normally. If, on the other hand, it has one or more radicals, steps 1 and 2 are repeated until obtaining an equation for radicals.
  4. Substitute the values obtained in the previous step into the initial equation and evaluate if they comply with the latter.

Sometimes, solutions, called strange roots, may appear that do not meet the original equation. Therefore, each of the solutions in the original equation must be verified and those that do not comply with it must be discarded.

Example 1: Solve 3√ (4x + 3) = 3

We raise both members to the cube:

[3√ (4x + 3)]3 = 3= 4x + 3 = 27

We subtract 3 from both members of the equation:

4x = 27 - 3

We divide by 4 both members of the equation:

x = 24/4

Possible solution:

x = 6

Substituting x = 6 into the original equation to evaluate whether it is a strange root or not, we find that 3√ (4x + 3) = 3, is correct. So:

Solution: x = 6

Example 2: Solve 3√ [2 + 5√ (x + 5)] = 2

We raise both members to the cube:

{ 3√ [2 + 5√ (x + 5)]}3 = 23

We remove the cubed radical:

2 + 5√ (x + 5) = 8

We subtract 2 from both members of the equation:

√ (x + 5) = 8 - 2

Let us observe that the unknown is under a radical sign:

√ (x + 5) = 6

Therefore, we square both members:

[√ (x + 5)]2 = 62

We remove the squared radical:

x + 5 = 36

We subtract 5 from both members of the equation:

x = 36 - 1

Possible solution:

x = 31

Substituting in the original equation to evaluate if it is a strange root or not, we check that 3√ {2 + 5√ [(31) + 5]} = 2, is correct. So:

Solution: x = 31

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

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 us observe that the unknown is 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)= to- 2ab + b:

x - 5 = 121 - 22x + x2

We obtain a quadratic equation of quadratic:

x2 - 23x + 126 = 0

Solving the equation, we obtain two possible solutions:

x = [- b ± √ (b2 - 4ac)] / 2a

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

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

x1 = [23 + √ (529 - 504)] / 2 = (23 + √25) / 2 = (23 + 5) / 2 = 28/2 = 14

x2 = [23 - √ (529 - 504)] / 2 = (23 - √25) / 2 = (23 - 5) / 2 = 18/2 = 9

By substituting x1 = 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:

x1 = 14 is a strange root

Now by substituting x2 = 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