What is simplification?

Simplification in math is to write a number into its constituent factor, or to write a number in a format to help perform mathematical operations.

A rational fraction of the format \(\dfrac{a}{\sqrt b} \) is simplified by removing the root symbol from the denominator by the process of rationalization.

Any of the arithmetic operations involving fractions can be performed conveniently only after rationalizing the denominator

In this mini-lesson, we will explore the topic of rationalizing the denominator, by finding answers to questions like what is the meaning of rationalizing, how to rationalize the denominator using conjugates, and check the solved examples, interactive questions.

**Lesson Plan**

**What Is the Meaning of Rationalizing?**

Rationalizing is the process of multiplying a surd with another similar surd, to result in a rational number.

The surd used to multiply is called the rationalizing factor (RF).

- To rationalize \(\sqrt x \) we need another \(\sqrt x \). \[\sqrt x \times \sqrt x = x \]
- To rationalize \(a + \sqrt b \) we need a rationalizing factor \(a - \sqrt b \). \[(a + \sqrt b) \times (a - \sqrt b) = (a)^2 - (\sqrt b)^2 = a^2 - b \]
- The rationalizing factor of \(2\sqrt3 \) is \(\sqrt3 \). \[2\sqrt3 \times \sqrt3 = 2\sqrt{3 \times 3} = 2 \times 3 = 6 \]

**How to Rationalize the Denominator Using Conjugates?**

Before we learn how to rationalize a denominator, we need to know about conjugates.

A conjugate is a similar surd but with a different sign.

The conjugate of \((\sqrt a + \sqrt b) \) is \( (\sqrt a - \sqrt b )\)

In the process of rationalizing a denominator, the conjugate is the rationalizing factor.

The process of rationalizing the denominator with its conjugate is as follows.

\[\begin{align} \frac{1}{\sqrt a + \sqrt b} &= \frac{1}{\sqrt a + \sqrt b} \times \frac{\sqrt a - \sqrt b}{\sqrt a - \sqrt b } \\ &= \frac{\sqrt a - \sqrt b}{(\sqrt a)^2 - (\sqrt b)^2} \\ &= \frac{\sqrt a - \sqrt b}{a - b} \end{align}\]

- Conjugate of \(\sqrt a + \sqrt b \) is \(\sqrt a - \sqrt b \)
- Rationalizing Factor (RF) of \(\sqrt x \) is \(\sqrt x \)
- Rationalizing factor of \(\sqrt x + \sqrt y \) is \(\sqrt x - \sqrt y \)
- The algebraic identity used in the process of rationalization is \((a + b)(a - b) = a^2 - b^2 \)

**How to Rationalize the Denominator Using Algebraic Identities?**

The algebraic formula used in the process of rationalization is \(a^2 - b^2 = (a + b)(a - b) \)

For rationalizing \((\sqrt a - \sqrt b )\), the rationalizing factor is \((\sqrt a + \sqrt b ) \). \[ (\sqrt a - \sqrt b ) \times (\sqrt a + \sqrt b ) =(\sqrt a)^2 - (\sqrt b)^2 = a - b \]

**Example**

Let us understand this with an example.

**Rationalize the denominator of **\(\dfrac{4}{\sqrt 11 - \sqrt 7} \)**.**

The conjugate of the denominator \(\sqrt11 - \sqrt 7\) is \(\sqrt11 + \sqrt7 \)

\[\begin{align}\dfrac{4}{\sqrt 11 - \sqrt 7} &=\dfrac{4}{\sqrt 11 - \sqrt 7} \times \dfrac{\sqrt11 + \sqrt7}{\sqrt 11 + \sqrt 7}\\ &= \dfrac{4(\sqrt11 + \sqrt7)}{(\sqrt 11)^2 - (\sqrt 7)^2} \\&=\dfrac{4(\sqrt11 + \sqrt7)}{11 - 7} \\ &=\dfrac{4(\sqrt11 + \sqrt7)}{4} \\&= \sqrt11 + \sqrt7\end{align} \]

**Solved Examples**

Example 1 |

Rationalize \(\dfrac{1}{\sqrt 7} \)

**Solution**

To rationalize the denominator \(\sqrt 7 \), we require another \(\sqrt 7 \)

\[\begin{align}\dfrac{1}{\sqrt 7} &= \dfrac{1}{\sqrt 7} \times \dfrac{\sqrt 7}{\sqrt 7} \\ &= \dfrac{\sqrt 7 }{\sqrt{ 7 \times 7}}\\ &=\dfrac{\sqrt 7 }{7}\end{align} \]

\(\therefore \text{The answer is} \dfrac{\sqrt 7 }{7} \) |

Example 2 |

Rationalize \(\dfrac{5}{\sqrt 5} \)

**Solution**

The rationalizing factor of \(\sqrt 5 \) is \(\sqrt 5 \)

\[\begin{align}\dfrac{5}{\sqrt 5} &= \dfrac{5}{\sqrt 5} \times \dfrac{\sqrt 5}{\sqrt 5} \\ &= \dfrac{5\sqrt5}{\sqrt{ 5 \times 5}} \\ &= \dfrac{5\sqrt 5}{ 5} \\ &=\sqrt5\end{align} \]

\(\therefore \text{The answer is} \sqrt5 \) |

Example 3 |

Rationalize the denominator of \(\dfrac{\sqrt 5}{(2 + \sqrt 3)} \)

**Solution**

The conjugate of the denominator \((2 + \sqrt 3 )\) is \((2 - \sqrt 3) \)

\[\begin{align}\dfrac{\sqrt 5}{(2 + \sqrt 3)} &=\dfrac{\sqrt 5}{(2 + \sqrt 3)} \times\dfrac{2 - \sqrt3}{2 - \sqrt3} \\ &=\dfrac{\sqrt5(2 - \sqrt3)}{2^2 - (\sqrt3)^2} \\ &=\dfrac{\sqrt5(2 - \sqrt3)}{4 - 3} \\ &= \dfrac{\sqrt5(2 - \sqrt3)}{1} \\ &=\sqrt5(2 - \sqrt3)\end{align} \]

\(\therefore \text{The answer is } \sqrt5(2 - \sqrt3)\) |

- Simplify \(\dfrac{5 + 4\sqrt 2}{3 – 2\sqrt 2} \) through rationalizing the denominator.
- Rationalize the denominator for the fraction \(\dfrac{2\sqrt3 + 4\sqrt 7}{4\sqrt3 – 2\sqrt 7} \).

**Interactive Questions on Rationalizing the Denominator**

**Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

The mini-lesson targeted the fascinating concept of rationalizing the denominator. The math journey around rationalizing the denominator starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

**About Cuemath**

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

**FAQs on Rationalize the Denominator**

### 1. How do you simplify radicals?

Radicals are the numbers of the form \(\sqrt x \).

To simplify radicals we need to multiply it with another radical.

### 2. What is a square root?

A square root is of the form \(\sqrt x \). To simplify a square root we need to have two similar factors within the square root.

\[ \sqrt 9 = \sqrt { 3 \times 3 } = 3 \]

### 3. What is a cube root?

A cube root is of the form \( \sqrt [3]a\). To simplify a cube root we need to have three similar factors within the cube root.

\[ \sqrt [3]{27} = \sqrt [3]{3 \times 3 \times 3} = 3\]

### 4. What is the conjugate of a complex number?

A complex number is of the form \((a + in) \), and its conjugate is \((a - ib) \).

### 5. How do you rationalize a denominator with two terms?

To rationalize the denominator with two terms, we multiply the numerator and denominator of the fraction with its conjugate.

To rationalize the denominator of \(\dfrac{1}{\sqrt3 - \sqrt2} \), we use its conjugate \((\sqrt3 + \sqrt2) \).

\[\begin{align} \dfrac{1}{\sqrt3 - \sqrt2} &= \dfrac{1}{\sqrt3 - \sqrt2} \times \dfrac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} \\&= \\ &=\dfrac{\sqrt3 + \sqrt2}{(\sqrt3)^2 - (\sqrt2)^2} \\ &= \dfrac{\sqrt3 + \sqrt2}{3 - 2} \\&=\dfrac{\sqrt3 + \sqrt2}{1} \\ &= \sqrt3 + \sqrt2\end{align} \]

### 6. How do you rationalize a square root?

To rationalize a \(\sqrt x \), we multiply it with the same root \(\sqrt x \). \[\sqrt5 \times \sqrt5 = \sqrt{5 \times 5} = 5 \]

### 7. How do you rationalize a surd?

The conjugate of \((\sqrt a + \sqrt b) \) is \((\sqrt a - \sqrt b) \). \[(\sqrt a + \sqrt b) \times (\sqrt a - \sqrt b) = (\sqrt a)^2 - (\sqrt b)^2 = a - b \]