Rationalizing Radical Expressions

What Rationalizing Changes

Rationalizing a denominator rewrites a fraction in an equivalent form whose denominator contains no radical. The value does not change because the numerator and denominator are multiplied by the same nonzero expression.

For a two-term denominator, the key identity is:

The expressions $u+v$ and $u-v$ are conjugates. Their product removes the middle terms.

One Radical in the Denominator

For $b>0$, multiply by the factor $\frac{\sqrt b}{\sqrt b}=1$:

The matching radical factor is a form of one. It is not called a conjugate because it has only one term.

A Sum or Difference of Two Radicals

Assume $a,b\geq0$ and $a\neq b$. Multiply by the conjugate of the denominator:

For example:

A Rational Term and a Radical Term

Assume $b\geq0$ and $a^2\neq b$. The conjugate again creates a difference of squares:

For example:

Simplifying Radical and Exponent Expressions

Assume the variables in the following exercises are positive. This assumption avoids the absolute-value cases that can appear when using fractional exponents.

1. Simplify:

   $$\left(\frac{8x^5y^{-4}}{16y^{-1/4}}\right)^{1/2}$$

2. Simplify:

   $$\left(5\sqrt{x^5}\right)\left(3\sqrt[3]x\right)$$

3. Simplify:

   $$\left(\frac{p^5q^{-10}}{p^5q^{-4}}\right)^{1/2}\left(\frac{p^{1/4}q^{-1/2}}{p^{-1/2}q^{-1/2}}\right)^{1/2}$$

Solutions for Simplifying Radicals

Each solution below rewrites the radicand as a product of powers before taking the root. Combine the numeric coefficients first, then handle each variable exponent on its own, and move any negative exponent into the denominator.

1. Combine the coefficient and each exponent before taking the square root:

   $$\left(\frac{8x^5y^{-4}}{16y^{-1/4}}\right)^{1/2}=\left(\frac12x^5y^{-15/4}\right)^{1/2}$$

   $$=2^{-1/2}x^{5/2}y^{-15/8}$$

   $$=\frac{x^{5/2}}{\sqrt2\,y^{15/8}}$$

2. Convert both radicals to fractional exponents:

   $$\left(5\sqrt{x^5}\right)\left(3\sqrt[3]x\right)=15x^{5/2}x^{1/3}$$

   $$=15x^{15/6+2/6}=15x^{17/6}$$

   $$=15x^2\sqrt[6]{x^5}$$

3. Simplify each parenthesis before multiplying:

   $$\left(q^{-6}\right)^{1/2}\left(p^{3/4}\right)^{1/2}=q^{-3}p^{3/8}$$

   $$=\frac{p^{3/8}}{q^3}$$

   $$=\frac{\sqrt[8]{p^3}}{q^3}$$

Rationalization Exercises

The exercises below ask you to remove a radical from the denominator. Multiply the numerator and the denominator by the same expression so the value of the fraction does not change, and choose that expression so the resulting denominator becomes rational.

1. For $b>0$, rationalize:

   $$\frac2{\sqrt{b^3}}$$

2. Rationalize:

   $$\frac2{\sqrt3+\sqrt5}$$

3. For $m\geq0$ and $m\neq n^2$, rationalize:

   $$\frac m{\sqrt m+n}$$

Solutions for Rationalizing Denominators

Each solution below multiplies by a form of one, either the radical itself or the conjugate. Watch how the chosen factor clears the root without changing the value of the fraction.

1. Because $\sqrt{b^3}\sqrt b=\sqrt{b^4}=b^2$ for $b>0$:

   $$\frac2{\sqrt{b^3}}\cdot\frac{\sqrt b}{\sqrt b}=\frac{2\sqrt b}{\sqrt{b^4}}$$

   $$=\frac{2\sqrt b}{b^2}$$

2. Multiply by the conjugate:

   $$\begin{aligned} &\frac2{\sqrt3+\sqrt5}\cdot\frac{\sqrt3-\sqrt5}{\sqrt3-\sqrt5} \\ &=\frac{2(\sqrt3-\sqrt5)}{3-5} \end{aligned}$$

   $$=\sqrt5-\sqrt3$$

3. The denominator's conjugate is $\sqrt m-n$:

   $$\frac m{\sqrt m+n}\cdot\frac{\sqrt m-n}{\sqrt m-n}=\frac{m(\sqrt m-n)}{m-n^2}$$

   $$=\frac{m\sqrt m-mn}{m-n^2}$$

How to Judge the Final Form

• The denominator should contain no radical.
• Every multiplying factor must equal one and must be defined under the stated conditions.
• An equivalent exact form is the goal. Rationalizing does not automatically make a decimal approximation more accurate.

These checks show why each step is valid and when the procedure can be used.