# Rationalize the Denominator

Go back to  'Irrational Numbers'

## Introduction:

Rationalizing the Denominator is a process to move a root (like a square root or cube root) from the bottom of a fraction to the top. We do it because it may help us to solve an equation easily.

Let us take an easy example, $$\frac{1}{{\sqrt 2 }}$$ has an irrational denominator. To make it rational, we will multiply numerator and denominator by $${\sqrt 2 }$$ as follows:

$\frac{1}{{\sqrt 2 }} = \frac{{1 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} = \frac{{\sqrt 2 }}{2}$

That is what we call Rationalizing the Denominator. Let's see how to rationalize other types of irrational expressions.

## Rationalizing the Denominator using conjugates:

Consider the irrational expression $$\frac{1}{{2 + \sqrt 3 }}$$. You have to express this in a form such that the denominator becomes a rational number. That is, you have to rationalize the denominator. You can do that by multiplying the numerator and the denominator of this expression by the conjugate of the denominator as follows:

\begin{align} \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} &= \frac{{2 - \sqrt 3 }}{{4 - 3}} \hfill \\ &= 2 - \sqrt 3 \hfill \\ \end{align}

Consider another example: $$\frac{{2 + \sqrt 7 }}{{2 - \sqrt 7 }}$$. We have to rationalize the denominator again, and so we multiply the numerator and the denominator by the conjugate of the denominator:

\begin{align} \frac{{2 + \sqrt 7 }}{{2 - \sqrt 7 }} &= \frac{{2 + \sqrt 7 }}{{2 - \sqrt 7 }} \times \frac{{2 + \sqrt 7 }}{{2 + \sqrt 7 }} \hfill \\ &= \frac{{4 + 7 + 4\sqrt 7 }}{{4 - 7}} \hfill \\ &= \frac{{11 + 4\sqrt 7 }}{{ - 3}} \hfill \\ \end{align}

### Using Algebraic Identities:

In carrying out rationalization of irrational expressions, we can make use of some general algebraic identities. For example, we already have used the following identity in the form of multiplying a mixed surd with its conjugate:

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

Here is another one:

$\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3}$

Let us take another problem of rationalizing the surd $$2 - \sqrt{7}$$. We make use of the second identity above. Here,

$\begin{gathered} a = 2,{\text{ }}b = \sqrt{7} \hfill \\ \Rightarrow {a^2} = 4,{\text{ }}ab = 2\sqrt{7},{\text{ }}{b^2} = \sqrt{{49}} \hfill \\ \end{gathered}$

We know that $$\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3}$$,

\begin{align} &\Rightarrow \left( {2 - \sqrt{7}} \right) \times \left( {4 + 2\sqrt{7} + \sqrt{{49}}} \right) \hfill \\ &= {\left( 2 \right)^3} - {\left( {\sqrt{7}} \right)^3} \hfill \\ &= 8 - 7 \hfill \\ &= 1 \hfill \\ \end{align}

## Solved Examples:

Example 1: Rewrite $$\frac{1}{{3 + \sqrt 2 - 3\sqrt 3 }}$$ by rationalizing the denominator:

Solution: Here, we have to rationalize the denominator. We can note that the denominator is a surd with three terms. Think: So what do we use as the multiplier?

We proceed as follows:

\begin{align} &\frac{1}{{\left( {3 + \sqrt 2 } \right) - 3\sqrt 3 }} \times \frac{{\left( {3 + \sqrt 2 } \right) + 3\sqrt 3 }}{{\left( {3 + \sqrt 2 } \right) + 3\sqrt 3 }} \hfill \\ = &\frac{{3 + \sqrt 2 + 3\sqrt 3 }}{{{{\left( {3 + \sqrt 2 } \right)}^2} - {{\left( {3\sqrt 3 } \right)}^2}}} \hfill \\ = &\frac{{3 + \sqrt 2 + 3\sqrt 3 }}{{9 + 2 + 6\sqrt 2 - 27}} \hfill \\ = &\frac{{3 + \sqrt 2 + 3 + \sqrt 3 }}{{ - 16 + 6\sqrt 2 }} \hfill \\ \end{align}

We note that the denominator is still irrational, which means that we have to carry out another rationalization step, where our multiplier will be the conjugate of the denominator:

\begin{align} &\frac{{3 + \sqrt 2 + 3\sqrt 3 }}{{ - 16 + 6\sqrt 2 }} \times \frac{{ - 16 - 6\sqrt 2 }}{{ - 16 - 6\sqrt 2 }} \hfill \\ = &\frac{{ - 48 - 18\sqrt 2 - 16\sqrt 2 - 12 - 48\sqrt 3 - 18\sqrt 6 }}{{{{\left( { - 16} \right)}^2} - {{\left( {6\sqrt 2 } \right)}^2}}} \hfill \\ = &\frac{{ - 60 - 34\sqrt 2 - 48\sqrt 3 - 18\sqrt 6 }}{{256 - 72}} \hfill \\ \end{align}

$= \boxed{ - \left( {\frac{{60 + 34\sqrt 2 + 48\sqrt 3 + 18\sqrt 6 }}{{184}}} \right)}$

Thus, using two rationalization steps, we have succeeded in rationalizing the denominator.

Example 2: Rationalize the denominator of the expression $$\frac{{2 - \sqrt{3}}}{{2 + \sqrt{3}}}$$.

Solution: In this case, we will use the following identity to rationalize the denominator: $$\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3}$$. We let

\begin{align} &a = 2,b = \sqrt{3}\\\Rightarrow &{a^2} = 4,ab = 2\sqrt{3},{b^2} = \sqrt{9} \end{align}

Now, we multiply the numerator and the denominator of the original expression by the appropriate multiplier:

\begin{align} &\frac{{2 - \sqrt{3}}}{{2 + \sqrt{3}}} \times \frac{{\left( {4 - 2\sqrt{3} + \sqrt{9}} \right)}}{{\left( {4 - 2\sqrt{3} + \sqrt{9}} \right)}} \hfill \\ = &\frac{{8 - 4\sqrt{3} + 2\sqrt{9} - 4\sqrt{3} + 2\sqrt{9} - \sqrt{{27}}}}{{{{\left( 2 \right)}^3} + {{\left( {\sqrt{3}} \right)}^3}}} \hfill \\ = &\frac{{8 - 8\sqrt{3} + 4\sqrt{9} - 3}}{{8 + 3}} \hfill \\ \end{align}

$\Rightarrow \boxed{\frac{{2 - \sqrt{3}}}{{2 + \sqrt{3}}} = \frac{{5 - 8\sqrt{3} + 4\sqrt{9}}}{{11}}}$

Example 3: Simplify the surd $$4\sqrt {12} - 6\sqrt {32} - 3\sqrt{{48}}$$ .

Solution: We have

$\begin{array}{l} 4\sqrt {12} = 4\sqrt {4 \times 3} = 8\sqrt 3 \\ 6\sqrt {32} = 6\sqrt {16 \times 2} = 24\sqrt 2 \\ 3\sqrt {48} = 3\sqrt {16 \times 3} =12\sqrt 3 \end{array}$

Thus, the given surd is equal to

$\boxed{\begin{array}{*{20}{l}} {8\sqrt 3 - 24\sqrt 2 - 12\sqrt 3 } \\ { = - 24\sqrt 2 - 12\sqrt 3 } \end{array}}$

Example 4: Suppose that $$x = \frac{{11}}{{4 - \sqrt 5 }}$$. Find the value of $${x^2} - 8x + 11$$ .

Solution: We rationalize the denominator of x:

\begin{align} x &= \frac{{11}}{{4 - \sqrt 5 }} \times \frac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}\\ &= \frac{{11\left( {4 + \sqrt 5 } \right)}}{{16 - 5}}\\ &= 4 + \sqrt 5 \\ \Rightarrow x - 4 &= \sqrt 5 \end{align}

Now, we square both the sides of this relation we have obtained:

\begin{align} {\left( {x - 4} \right)^2} &= 5 \hfill \\ \Rightarrow {x^2} - 8x + 16 &= 5 \hfill \\ \end{align}

$\Rightarrow \boxed{{x^2} - 8x + 11 = 0}$

Example 5: Suppose that a and b are rational numbers such that

$\frac{{3 + 2\sqrt 3 }}{{5 - 2\sqrt 3 }} = a + b\sqrt 3$

Find the values of a and b.

Solution: We rationalize the denominator of the left-hand side (LHS):

\begin{align} {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} &= \frac{{3 + 2\sqrt 3 }}{{5 - 2\sqrt 3 }} \times \frac{{5 + 2\sqrt 3 }}{{5 + 2\sqrt 3 }} \hfill \\ &= \frac{{15 + 6\sqrt 3 + 10\sqrt 3 + 12}}{{{{\left( 5 \right)}^2} - {{\left( {2\sqrt 3 } \right)}^2}}} \hfill \\ &= \frac{{27 + 16\sqrt 3 }}{{25 - 12}} \hfill \\ &= \frac{{27}}{{13}} + \frac{{16}}{{13}}\sqrt 3 \hfill \\ \end{align}

Comparing this with the right hand side of the original relation, we have $$\boxed{a = \frac{{27}}{{13}}}$$ and $$\boxed{b = \frac{{16}}{{13}}}$$. Challenge: Simplify the following expression:

$\frac{1}{{\sqrt 3 - \sqrt 4 }} + \frac{1}{{\sqrt 3 + \sqrt 4 }}$

⚡Tip: Take LCM and then apply property, $$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$$.

Numbers and Number Systems
Numbers and Number Systems
grade 9 | Questions Set 2
Numbers and Number Systems