Simplifying Radicals

We will use the following conventions for simplifying expressions involving radicals:

1. Given the expression $a^{\frac {b}{c}}$, write this as ${\sqrt[{c}]{a^{b}}}$
2. No fractions under the radical sign
3. No radicals in the denominator
4. The radicand has no exponentiated factors with exponent greater than or equal to the index of the radical

Example: Simplify the expression $\left({\frac {1}{8}}\right)^{\frac {1}{2}}$ Using convention 1, we rewrite the given expression as

$\left({\frac {1}{8}}\right)^{\frac {1}{2}}={\sqrt[{2}]{\left({\frac {1}{8}}\right)^{1}}}={\sqrt {\frac {1}{8}}}$

The expression now violates convention 2. To get rid of the fraction in the radical, apply the rule $\left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}$ and simplify the result:

${\sqrt {\frac {1}{8}}}={\frac {\sqrt {1}}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}$

The resulting expression violates convention 3. To get rid of the radical in the denominator, multiply by ${\frac {\sqrt {8}}{\sqrt {8}}}$:

${\frac {1}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}\cdot {\frac {\sqrt {8}}{\sqrt {8}}}={\frac {\sqrt {8}}{8}}$

Notice that $8=2^{3}$. Since the index of the radical is 2, our expression violates convention 4. We can reduce the exponent of the expression under the radical as follows:

${\frac {\sqrt {8}}{8}}={\frac {\sqrt {2^{3}}}{8}}={\frac {\sqrt {2^{2}\cdot 2}}{8}}={\frac {2\cdot {\sqrt {2}}}{8}}={\frac {\sqrt {2}}{4}}$

The Conjugate of a Radical Expression

The conjugate of the two term expression $a+b$ is $a-b$ (likewise, the conjugate of $a-b$ is $a+b$). The product of a two term expression and its conjugate, $a+b$ and $a-b$, is $(a+b)(a-b)=a^{2}+ba-ba-b^{2}=a^{2}-b^{2}$. This property is useful for getting rid of square roots in two term expressions.

For example, consider the limit

$\lim _{h\to 0}{\frac {{\sqrt {3+h}}-{\sqrt {3}}}{h}}$

If we plug in 0 for $h$, we get the indeterminate form $0/0$. However, using the conjugate of the numerator (${\sqrt {3+h}}+{\sqrt {3}}$) and multiplying by a clever form of 1 (the conjugate divided by itself), we can rewrite this limit in a way that allows us to evaluate it.

$\lim _{h\to 0}{\frac {{\sqrt {3+h}}-{\sqrt {3}}}{h}}=\lim _{h\to 0}{\frac {({\sqrt {3+h}}-{\sqrt {3}})}{h}}\cdot {\frac {({\sqrt {3+h}}+{\sqrt {3}})}{({\sqrt {3+h}}+{\sqrt {3}})}}$

$=\lim _{h\to 0}{\frac {({\sqrt {3+h}})^{2}-({\sqrt {3}})^{2}}{h({\sqrt {3+h}}+{\sqrt {3}})}}$

$=\lim _{h\to 0}{\frac {((3+h)-3}{h({\sqrt {3+h}}+{\sqrt {3}})}}$

$=\lim _{h\to 0}{\frac {h}{h({\sqrt {3+h}}+{\sqrt {3}})}}$

$=\lim _{h\to 0}{\frac {1}{{\sqrt {3+h}}+{\sqrt {3}}}}$

Now, setting $h=0$ will not result in an indeterminate form, so we can evaluate the limit.

$={\frac {1}{{\sqrt {3+0}}+{\sqrt {3}}}}={\frac {1}{2{\sqrt {3}}}}$

Resources

• Introduction to Radicals, Lumen Learning: Boundless Algebra
• Simplifying Square Roots, Khan Academy
• Useful Radical/Root Rules for Simplification, Mathwords.com
• Example Problems of Simplifying Radicals, LibreTexts