Rationalizing the Denominator


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Rationalizing the Denominator

It is often preferable to rewrite fractions with radicals in the denominator as equivalent fractions with radicals in the numerator only.

\(\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\)

 

 

Example 15

Rationalize the denominator of \(\dfrac{4}{\sqrt{3}}\).

 

Example 15 Continued

Rationalize the denominator of \(\dfrac{4}{\sqrt{3}}\).

 

 

Example 16

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

 

 

Example 17

Rationalize the denominator of \(\dfrac{2}{5\sqrt{6}}\).

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Check Your Understanding 4


Rationalize the denominator of \(\dfrac{a}{c\sqrt{b}}\).

Enter \(\sqrt{a}\) as "sqrt\((a)\)" and \(\dfrac{a}{b}\) as "\(a/b\)".

There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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How Did I Do?Try Another

So far, all of the examples we have seen have a single term in their numerator.  This won't always be the case; however, the process is the same even if we have more than one term in the numerator.

Example 18

Rationalize the denominator of \(\dfrac{10-\sqrt{5}}{\sqrt{5}}\).

Solution

Approach 1

Rewrite this fraction without \(\sqrt{5}\) in the denominator.

\(\dfrac{10-\sqrt{5}}{\sqrt{5}}\;\)

\(=\dfrac{10-\sqrt{5}}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\)

  • Multiply numerator and denominator by \(\sqrt{5}\) .

 

\(=\dfrac{(10-\sqrt{5})\times\sqrt{5}}{\sqrt{5}\times\sqrt{5}}\)

  • Expand and simplify both the numerator and denominator.

 

\(=\dfrac{10\sqrt{5}-\sqrt{5\times5}}{\sqrt{5\times5}}\)

 

 

\(=\dfrac{10\sqrt{5}-\sqrt{25}}{\sqrt{25}}\)

 

 

\(=\dfrac{10\sqrt{5}-5}{5}\)

  • Common factor \(5\) from both terms in the numerator.

 

\(=\dfrac{5(2\sqrt{5}-1)}{5}\)

  • Reduce the fraction.

 

\(=\dfrac{5}{5}(2\sqrt{5}-1)\)

 

 

\(=2\sqrt{5}-1\)

 

Approach 2

You may have recoginzed that \(\sqrt{x}\times \sqrt{x}=x\) for \(x \geq 0\). If you recognized this, the same steps can be followed as above but the solution can be shortened to:

\(\dfrac{10-\sqrt{5}}{\sqrt{5}}\;\)

\(=\dfrac{10-\sqrt{5}}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}\)

  • Multiply numerator and denominator by \(\sqrt{5}\).

 

\(=\dfrac{10\sqrt{5}-\sqrt{5}\times \sqrt{5}}{5}\)

  • Expand and simplify both the numerator and denominator.

 

\(=\dfrac{10\sqrt{5}-5}{5}\)

 

 

\(=\dfrac{5(2\sqrt{5}-1)}{5}\)

  • Common factor \(5\) from both terms in the numerator.

 

\(=\dfrac{5}{5}(2\sqrt{5}-1)\)

  • Reduce the fraction.

 

\(=2\sqrt{5}-1\)

 

Before we continue our work with radicals, let's review expanding binomials using difference of squares. 

Recall

Difference of Squares:

\((a+b)(a-b)=a^2-b^2\)

For instance, 

\((x+2)(x-2)\)

\(=x^2-2x+2x-4\)

 

\(=x^2-4\)

Example 19

Simplify \((\sqrt{2}+\sqrt{7})(\sqrt{2}-\sqrt{7})\).

Solution

\(\begin{align*} (\sqrt{2}+\sqrt{7})(\sqrt{2}-\sqrt{7}) &=\sqrt{2}(\sqrt{2})+\sqrt{2}(-\sqrt{7})+\sqrt{7}(\sqrt{2})+\sqrt{7}(-\sqrt{7}) \\ &=\sqrt{4}-\sqrt{14}+\sqrt{14}-\sqrt{49} \\ &=2 +0 -7 \\ &=-5 \end{align*}\)

Notice that we can evaluate two of the radicals and the remaining radicals have opposite values of one another.  This is not a coincidence!

\((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b\) for \(a\),\(b \geq 0\)

 

The Conjugate

Let's take a look at the conjugate for radical binomials. The conjugate for a radical expression \(a+b\sqrt{c}\)  has the same first and last term, but contains the opposite sign between these terms.

The conjugate of \(a+b\sqrt{c}\)  is \(a-b\sqrt{c}\) , for rational numbers \(a,~b\) and \(c\).

We can use the conjugate to rationalize the denominator of radical expressions, that is to rewrite a radical expression as an equivalent fraction with radicals in the numerator only.


Example 20

Rationalize the denominator of \(\dfrac{-3}{1+\sqrt{2}}\).

The conjugate of the radical expression \(1+\sqrt{2}\)  is \(1-\sqrt{2}\).

Approach 1 — Expand the denominator

\(\dfrac{-3}{1+\sqrt{2}}\;\)

\(=\dfrac{-3}{1+\sqrt{2}}\times\dfrac{1-\sqrt{2}}{1-\sqrt{2}}\)

  • Multiply numerator and denominator by \(1-\sqrt{2}\).

 

\(=\dfrac{-3\times(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}\)

  • Do not expand the numerator.

 

\(= \dfrac{-3\times(1-\sqrt{2})}{1-\sqrt{2}+\sqrt{2}-\sqrt{4}}\)

  • Expand and simplify the denominator.

 

\(= \dfrac{-3\times(1-\sqrt{2})}{1-2}\)

 

 

\(= \dfrac{-3\times(1-\sqrt{2})}{-1}\)

 

 

\(= \dfrac{-3}{-1}(1-\sqrt{2})\)

  • Reduce the fraction.

 

\(=3(1-\sqrt{2})\)

 

Approach 2 — Use difference of squares, \((a+b)(a-b)=a^2-b^2\), to simplify the denominator

\(\dfrac{-3}{1+\sqrt{2}}\)

\(=\dfrac{-3}{1+\sqrt{2}}\times\dfrac{1-\sqrt{2}}{1-\sqrt{2}}\)

  • Multiply numerator and denominator by \(1-\sqrt{2}\).

 

\(=\dfrac{-3\times(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}\)

  • Do not expand the numerator.

 

\(= \dfrac{-3(1-\sqrt{2})}{(1)^2-(\sqrt{2})^2}\)

  • Expand and simplify the denominator (using difference of squares).

 

\(= \dfrac{-3(1-\sqrt{2})}{1-2}\)

  • Reduce the fraction.

 

\(= \dfrac{-3(1-\sqrt{2})}{-1}\)

 

 

\(= \dfrac{-3}{-1}(1-\sqrt{2})\)

 

 

\(=3(1-\sqrt{2})\)

 

It is more efficient to simplify the expression by using the difference of squares. This method will be used moving forward.

Example 21

Rationalize the denominator of \(\dfrac{1+\sqrt{3}}{2-\sqrt{5}}\).

Solution

The conjugate of \(2-\sqrt{5}\) is \(2+\sqrt{5}\).

\(\dfrac{1+\sqrt{3}}{2-\sqrt{5}}\;\)

\(=\dfrac{1+\sqrt{3}}{2-\sqrt{5}}\times\dfrac{2+\sqrt{5}}{2+\sqrt{5}}\)

  • Multiply the numerator and denominator by \(2+\sqrt{5}\).

 

\(=\dfrac{(1+\sqrt{3})(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})}\)

  • Expand and simplify the denominator (using difference of squares).

 

\(=\dfrac{(1+\sqrt{3})(2+\sqrt{5})}{(2)^2-(\sqrt{5})^2}\)

 

 

\(=\dfrac{(1+\sqrt{3})(2+\sqrt{5})}{4-5}\)

 

 

\(=\dfrac{(1+\sqrt{3})(2+\sqrt{5})}{-1}\)

  • Remember that dividing by \(-1\) is equivalent to multilpying by \(-1\). 

 

\(=-1((1+\sqrt{3})(2+\sqrt{5}))\)

  • Expand using the distributive property.

 

\(=-1(2+\sqrt{5}+2\sqrt{3}+\sqrt{15})\)

 

 

\(=-2-\sqrt{5}-2\sqrt{3}-\sqrt{15}\)

 

Example 22

Rationalize the denominator of \(\dfrac{a}{\sqrt{a}-\sqrt{b}}\).

Solution

The denominator in this example looks a little different from those that we have seen in previous examples.  We can rationalize the denominator by multiplying numerator and denominator by \(\sqrt{a}+\sqrt{b}\).

\(\dfrac{a}{\sqrt{a}-\sqrt{b}}\;\)

\(=\dfrac{a}{\sqrt{a}-\sqrt{b}}\times \dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

  • Multiply the numerator and denominator by \(\sqrt{a}+\sqrt{b}\).

 

\(=\dfrac{a(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}\)

  • Expand and simplify the denominator (using difference of squares).

 

\(=\dfrac{a(\sqrt{a}+\sqrt{b})}{(\sqrt{a})^2-(\sqrt{b})^2}\)

 

 

\(=\dfrac{a(\sqrt{a}+\sqrt{b})}{a-b}\)

 

 

\(=\dfrac{a(\sqrt{a}+\sqrt{b})}{a-b}\)

 

Note that we could have also multiplied numerator and denominator by \(-\sqrt{a}-\sqrt{b}\) to rationalize the denominator.


Check Your Understanding 5


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Try This Revisited

Each expression is equivalent to one other expression in the following list.

Identify the pairs of equivalent expressions.

  1. \(\sqrt{2}\left(\sqrt{2}+\sqrt{32}\right)\)
  2. \((5+\sqrt{3})(4-\sqrt{2})\)
  3. \(\sqrt{48}+20-\sqrt{6}-\sqrt{50}\)
  4. \(1-\sqrt{18}+\sqrt{2}-\sqrt{16}\)
  5. \(\sqrt{27}+\sqrt{100}-\sqrt{3}-\sqrt{12}\)
  6. \(\dfrac{1+\sqrt{2}}{1-\sqrt{2}}\)

Begin by simplifying each expression. Then compare simplified expressions for equivalency.

Solution — Part A

Expand and simplify.

\[\begin{align*} \sqrt{2}\left(\sqrt{2}+\sqrt{32}\right)&=\sqrt{2}\times \sqrt{2}+\sqrt{2}\times \sqrt{32}\\ &=\sqrt{4}+\sqrt{64}\\ &=2+8\\ &=10 \end{align*}\]

Solution — Part B

Use the Distributive Property to expand and simplify.

\[\begin{align*} (5+\sqrt{3})(4-\sqrt{2})&=20+5(-\sqrt{2})+\sqrt{3}(4)+\sqrt{3}(-\sqrt{2})\\ &=20-5\sqrt{2}+4\sqrt{3}-\sqrt{6} \end{align*}\]

Solution — Part C

Rewrite each term as a mixed radical, then collect like terms.

\[\begin{align*} \sqrt{48}+20-\sqrt{6}-\sqrt{50}&=\sqrt{16\times3}+20-\sqrt{6}-\sqrt{25\times2}\\ &=\sqrt{16}\sqrt{3}+20-\sqrt{6}-\sqrt{25}\sqrt{2}\\ &=4\sqrt{3}+20-\sqrt{6}-5\sqrt{2}\\ \end{align*}\]

Solution — Part D

Rewrite each term as a mixed radical, then collect like terms.

\[\begin{align*} 1-\sqrt{18}+\sqrt{2}-\sqrt{16}&=1-\sqrt{9\times2}+\sqrt{2}-4\\ &=1-\sqrt{9}\sqrt{2}+\sqrt{2}-4\\ &=1-3\sqrt{2}+\sqrt{2}-4\\ &=-3-2\sqrt{2} \end{align*}\]

Solution — Part E

Rewrite each term as a mixed radical, then collect like terms.

\[\begin{align*} \sqrt{27}+\sqrt{100}-\sqrt{3}-\sqrt{12}&=\sqrt{9\times3}+10-\sqrt{3}-\sqrt{4\times3}\\ &=\sqrt{9}\sqrt{3}+10-\sqrt{3}-\sqrt{4}\sqrt{3}\\ &=3\sqrt{3}+10-\sqrt{3}-2\sqrt{3}\\ &=10 \end{align*}\]

Solution — Part F

Rationalize the denominator.

\[\begin{align*} \dfrac{1+\sqrt{2}}{1-\sqrt{2}}&=\dfrac{1+\sqrt{2}}{1-\sqrt{2}}\times\dfrac{1+\sqrt{2}}{1+\sqrt{2}}\\ &=\dfrac{(1+\sqrt{2})(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}\\ &=\dfrac{(1+\sqrt{2})^2}{1-2}\\ &=\dfrac{(1+\sqrt{2})^2}{-1}\\ &=-(1+\sqrt{2})^2 \end{align*}\]

We can see that:

  • Parts a) and e) both simplify to \(10\), so \(\sqrt{2}\left(\sqrt{2}+\sqrt{32}\right)\) and \(\sqrt{27}+\sqrt{100}-\sqrt{3}-\sqrt{12}\) are equivalent expressions.
  • Parts b) and c) both simplify to \(4\sqrt{3}+20-\sqrt{6}-5\sqrt{2}\), so \((5+\sqrt{3})(4-\sqrt{2})\) and \(\sqrt{48}+20-\sqrt{6}-\sqrt{50}\) are equivalent expressions.

  • Part d) simplifies to \(-3-2\sqrt{2}\) and part f) simplifies to \(-(1+\sqrt{2})^2\). They do not appear the same but they are also not in the same form. Expand part f), and then compare the simplified expressions.

    \[\begin{align*} -(1+\sqrt{2})^2&=-(1+\sqrt{2})(1+\sqrt{2})\\ &=-\left(1+\sqrt{2}+\sqrt{2}+\sqrt{4} \right)\\ &=-\left(1+2\sqrt{2}+2 \right)\\ &=-\left(3+2\sqrt{2} \right)\\ &=-3-2\sqrt{2} \end{align*}\]
  • Parts d) and f) both simplify to \(-3-2\sqrt{2}\), so \(1-\sqrt{18}+\sqrt{2}-\sqrt{16}\) and \(\dfrac{1+\sqrt{2}}{1-\sqrt{2}}\) are equivalent expressions.