Radical is used to represent a Square Root of a number. For Example: 5 is square root of number 25. We will denote a radical by symbol ($\sqrt{}$). This is called radical or radix. Number which is put under radical symbol is called a radicand. When a number is written under the radical sign then it is called as radical form. Solving equations is the process of finding the values of the variables or unknown quantities that are satisfied by the condition defined in the equation. An equation is described as the combination of the variables, constants and operations. In this section will discuss about solution of radical equations and how to solve:
  • Radical Equations with one Radical 
  • Radical Equations with Two Radicals 
  • Radical Equations with Rational Exponents and many more

What are Radical Equations

Radical expressions are algebraic expressions containing the radical sign, commonly known as root sign. Radicals denote the inverse operation of exponentiation and also equivalently represented by using fractional exponents. For being able to solve radical equations, it is imperative that we first get rid of the radical. Just like how a polynomial equation can be linear, quadratic, cubic, etc, a radical equation can also be a square root equation, a cubed root equation, a fourth root equation etc. 
Radical equations are equations with radicals where the radicand contain the variable.

Examples:

 Sl No. Equation  Explanation
 1 $\sqrt{x + 4} - 2$ = 2  Variable $x$ is present in the radicand x + 4 
 2 $\sqrt{2x + 2}$ = x + 6  Variable $x$ is present in the radicand 2x + 2 as well as outside the radical.
 3 $\sqrt{2x + 5} + \sqrt{3x - 2}$ = 4    The equation contains two radicals.
  .

Steps to Solve Radical Equations

In general the steps for solving radical equations can be summarized as follows:

1) First we need to isolate the radical expression. If there are two radical expressions, then we need to have them both on different sides of the equal to sign. Note that it cannot be solve with two radical expressions on same side, but it is less cumbersome the other way.

2) Next we square both the sides. This is the main step in solving radical equations. For example, if we have a = b, then on squaring both the sides it becomes $a^{2}$ = $b^{2}$.
3) As a result of above step, the radical can be removed. Now that we have got rid of the radical, we can solve the equation for the unknown variable like any other linear or quadratic equation.

4) The last and most important step. We need to plug back the answers into the original radical equation and check if all the answers that we got actually work or not. It is possible that some of the answers do not work. Such solutions are called ‘extraneous’ solutions.

Solving Radical Equations with One Radical

Radical functions contain functions involving roots. A radical function contains a radical expression with the independent variable in the radicand. An example of radical function is y = $\sqrt{x}$. 

Domain of the function should not have a negative value under the square root sign. Otherwise, it will lead to complex number.

Let us solve 5 = $\sqrt{3x}$ + 12 
Isolate the radical by subtracting 12 from both the sides.
5 - 12 = $\sqrt{3x}$ + 12 - 12

-7 = $\sqrt{3x}$

Squaring on both sides, we get
49 = 3x
Simplify, 
x = 49/3 = 16.33

Solving Radical Equations with Two Radicals

Sometimes, it is required to square the radical equation twice to get rid of the square roots. Let us see with the help of an example how to solve these type of equations.

For example, consider the equation $\sqrt{x + 1}$ - $\sqrt{2x - 2}$ = $0$

$\sqrt{x + 1}$ = $\sqrt{2x - 2}$                         (Isolate the radicals)

$(\sqrt{x + 1})^{2}$ = $(\sqrt{2x - 2})^{2}$           (Square the equation)

$x + 1$ = $2x - 2$ $\Rightarrow\ x$ = $3$

And, this solution also verifies in the original equation.

Examples

Example 1: Solve  $\sqrt{(x + 5)}$ = $3$

Solution: Square both the sides.

$\sqrt{(x + 5)^{2}}$ = $3^{2}$

or x + 5 = 9

Subtracting 5 from both the side

x + 5 - 5 = 9 - 5

x = 4.  Answer!

Verify your answer: Check the answer by plugging it into the original equation. 

$\sqrt{(x + 5)}$ = 3

$\sqrt{(4 + 5)}$ = 3

$\sqrt{9}$ = 3

3 = 3

Yes! It works! So the answer that we found is correct.

Example 2: Solve for x and check for extraneous solutions: $\sqrt{2} - \sqrt{x + 6} = -\sqrt{x}$

Solution: $\sqrt{2} - \sqrt{x + 6} = -\sqrt{x}$

$[\sqrt{2} - \sqrt{x + 6}]^{2} = (-\sqrt{x})^{2}$                 (Square the Equation)

2 - $2\sqrt{2(x + 6)}$ + x + 6 = x                                    (Expand the left side)

$-2\sqrt{2x + 12}$ = -8                                                   (Radical isolated)

$\sqrt{2x + 12}$ = 4
After simplifying, we did not get rid of radical. So repeat the steps to isolate x.

$[\sqrt{2x + 12}]^{2}$ = 42                                               (Square the equation)

2x + 12 = 16
x = 2

Check the solution:
$\sqrt{2} - \sqrt{2 + 6} = -\sqrt{2}$
$\sqrt{2} - \sqrt{8} = -\sqrt{2}$
$\sqrt{2} - 3\sqrt{2} = -\sqrt{2}$
$-\sqrt{2} = -\sqrt{2}$
Thus, x = 2 is the solution.

Practice Problems

Solve the below problems:
1. $\sqrt{x + 4} - 2$ = 2
2. $\sqrt{2x + 5} + \sqrt{3x - 2}$ = 4