Solving equations is a major part of math problems. Students come across various types of equations while solving the problems. Some of these problems involve solving radical equations.

The process of solving a radical equation is a bit different from the others. Letâ€™s understand how these types of equations are solved.

## What is a Radical in Math?

In math, a radical is the opposite of an exponent that is represented with a symbol ‘$\sqrt {}$’ and is also known as the root. The term radical is derived from the Latin word Radix which means root. It can either be a square root or a cube root or $n^{th}$ root. The number before the symbol or radical is considered to be an index number or degree.

Examples of radicals are $\sqrt{5}$, $\sqrt[3]{2}$, $\sqrt[4]{7}$, $\sqrt[n]{a}$.

The number is written before the radical is known as the index number or a degree. This number helps in telling us how many times the number would be multiplied by itself to equal the radicand.

This is considered to be the opposite of an exponent just like addition being the opposite of subtraction and division being the opposite of multiplication.

For example: $\sqrt[3]{125}=5$ as $5 \times 5 \times 5 = 125$.

## What is a Radical Equation?

A radical equation is an equation in which a variable is under a radical.

The variable under radical can be on

- one side of the equation
- one radical variable
- more than one radical variable

- both sides of the equation

Example of equation with radical variable on one side of the equation is $\sqrt{x^2 + 2x – 3} = 3$

Example of equation with radical variable on both sides of the equation is $ \sqrt{2x^2 – 3x + 5} = \sqrt{x^2 + x – 1} + 2$

## How to Solve Radical Equations?

To solve a radical equation:

**Step 1:** Isolate the radical expression involving the variable. If more than one radical expression involves the variable, then isolate one of them.

**Step 2:** Raise both sides of the equation to the index of the radical.

**Step 3:** If there is still a radical equation, repeat steps 1 and 2; otherwise, solve the resulting equation and check the answer in the original equation.

By raising both sides of an equation to power, some solutions may have been introduced that do not make the original equation true. These solutions are called **extraneous solutions**.

## How To Solve Square Root Equations?

Solving equations with the radical variables on one side. To understand the process, letâ€™s consider the following example:

**Case 1: One side of the equation has one radical variable**

Solve $\sqrt {3x^2 + 10x} – 5 = 0$

Take 5 to the other side. (Separate radical and non-radical variables)

$\sqrt {3x^2 + 10x} = 5$

Square both sides

${\sqrt {3x^2 + 10x}}^2 = 5^{2}$

$=>3x^2 + 10x = 25$

$=>3x^2 + 10x – 25 = 0$

This is the quadratic equation and now it can be solved using any of the methods (Splitting the middle term, using the quadratic formula or completing the square method)

Letâ€™s solve it by using the quadratic formula.

$a = 3, b = 10, c = -25$

By quadratic formula $x = \frac {-b \pm \sqrt{b^2 – 4ac}}{2a}$

$=> x = \frac {-10 \pm \sqrt{10^2 – 4 \times 3 \times \left( -25 \right) }}{2 \times 3}$

$=> x = \frac {-10 \pm \sqrt{100 + 300 }}{6}$

$=> x = \frac {-10 \pm \sqrt{400}}{6}$

$=> x = \frac {-10 \pm 20}{6}$

$=> x = \frac {-10 + 20}{6}$ or $x = \frac {-10 – 20}{6}$

$=> x = \frac {10}{6}$ or $x = \frac {-30}{6}$

$=> x = \frac {5}{3}$ or $x = -5$

Now, check the results

When x = \frac {5}{3}$

$\sqrt {3 \times {\frac {5}{3}}^2 + 10 \times \frac {5}{3}} – 5 = 0$

$=>\sqrt {3 \times \frac {25}{9} + \frac {50}{3}} – 5 = 0$

$=>\sqrt {\frac {75}{9} + \frac {50}{3}} – 5 = 0$

$=>\sqrt {\frac {25}{3} + \frac {50}{3}} – 5 = 0$

$=>\sqrt {\frac {75}{3}} – 5 = 0$

$=>\sqrt {25} – 5 = 0$

$=>5 – 5 = 0$ $=>0 = 0$ is true

When $x = -5$

$\sqrt {3 \times \left( -5 \right)^{2} + 10 \times \left( -5 \right)} – 5 = 0$

$=> \sqrt {3 \times 25 – 50} – 5 = 0$

$=> \sqrt {75 – 50} – 5 = 0$

$=> \sqrt {25} – 5 = 0$

$=> 5 – 5 = 0$ $=> 0 = 0$ is true

**Case 2: One side of the equation has more than one radical variable**

$ \sqrt {3x – 5} + \sqrt {x – 1} = 2$

Isolate the radical variables

$ \sqrt {3x – 5} = 2 – \sqrt {x – 1} $

Square both sides of the equation

$ \left( \sqrt {3x – 5} \right)^{2} = \left( 2 – \sqrt {x – 1} \right)^{2}$

$ => 3x – 5 = \left(2 – \sqrt {x – 1} \right) \times \left(2 – \sqrt {x – 1} \right) $

$ => 3x – 5 = 4 – 2 \sqrt{x – 1} – 2 \sqrt{x – 1} – x – 1$

$ => 3x – 5 = 3 – 4 \sqrt{x – 1} + x$

This is still a radical equation. Again isolate the radical expression.

$ => 3x – 5 – 3 – x = – 4 \sqrt{x – 1}$

$ => 2x – 8 = – 4 \sqrt{x – 1}$

$ => 2 \left( x – 4 \right) = – 4 \sqrt{x – 1}$

$ => x – 4 = – 2 \sqrt{x – 1}$

Square both sides

$ \left( x – 4 \right)^{2} = \left( – 2 \sqrt{x – 1} \right)^{2}$

$ => x^{2} – 8x + 16 = 4 \left( x – 1 \right) $

$ => x^{2} – 8x + 16 = 4x – 4 $

$ => x^{2} – 8x + 16 – 4x + 4 = 0$

$ => x^{2} – 12x + 20 = 0 $

Letâ€™s solve it by factorization method (splitting the middle term)

$ x^{2} – 2x – 10x + 20 = 0$

$=> x \left( x – 2 \right) – 10 \left( x – 2 \right) = 0$

$=> \left( x – 2 \right) \left( x – 10 \right) = 0$

$=> x – 2 = 0 $ or $ x – 10 = 0 $

$=> x = 2 $ or $x = 10$

Now, check the results

When $=> x = 2 $

$ \sqrt {3 \times 2 – 5} + \sqrt {2 – 1} = 2$

$ =>\sqrt {6 – 5} + \sqrt {1} = 2$

$ =>\sqrt {1} + \sqrt {1} = 2$

$ =>1 + 1 = 2$

$ => 2 = 2 $ is true

When $=> x = 10 $

$ \sqrt {3 \times 10 – 5} + \sqrt {10 – 1} = 2$

$=> \sqrt {30 – 5} + \sqrt {9} = 2$

$=> \sqrt {25} + \sqrt {9} = 2$

$=> 5 + 3 = 2$

$=> 8 = 2$ is false

So, $ \sqrt {3x – 5} + \sqrt {x – 1} = 2$ has only one solution and that is $x = 2$.

## Conclusion

The process of solving a radical equation involves two steps viz., the first removing the radical part and the second solving the simplified equation. Whenever you solve a radical equation keep repeating the first part, till all the radical parts of the equation are eliminated.

## Practice Problems

Solve the following equations:

- $2x = \sqrt {x + 3}$
- $\sqrt {33 – 2x} = x + 1$
- $7 = \sqrt {39 + 3x} – x$
- $x = 1 + \sqrt {2x – 2} $
- $1 + \sqrt {1 – x} = \sqrt {2x + 4}$
- $\sqrt {5x – 4} – 9 = 0$
- $ \sqrt {3x + 2} – 5 = 0$
- $ \sqrt {10x + 1} – 2 = 0$
- $ \sqrt {9k – 2} + 1 = 0$
- $ \sqrt {7x – 3} + 2 = 0$
- $ \sqrt {x – 1} + 1 = p$
- $ \sqrt [3] {4x – 3} + 8 = 5$
- $ \sqrt [3] {6x – 10} + 1 = -3$
- $ \sqrt [4] {3x – 2} + 3 = 5$
- $ \sqrt [4] {4x – 8} + 5 = 7$