Simplifying Radicals

Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. Generally speaking, it is the process of simplifying expressions applied to radicals.

A radical is a number that has a fraction as its exponent:

$$\sqrt[n]{x^m} = x ^ { m/n }.$$

Since they are exponents, radicals can be simplified using rules of exponents.

The square root of a positive integer that is not a perfect square is always an irrational number. The decimal representation of such a number loses precision when it is rounded, and it is time-consuming to compute without the aid of a calculator. Instead of using decimal representation, the standard way to write such a number is to use simplified radical form, which involves writing the radical with no perfect squares as factors of the number under the root symbol.

Let $a$ be a positive non-perfect square integer.

The simplified radical form of the square root of $a$ is

$$\sqrt{a}=b\sqrt{c}.$$

In this form $\sqrt{a}=b\sqrt{c}$, both $b$ and $c$ are positive integers, and $c$ contains no perfect square factors other than $1$.

The process for putting a square root into simplified radical form involves finding perfect square factors and then applying the identity $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$, which allows us to take the root of the perfect square factors.

Simplify $\sqrt{12}$.

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Since $12 = 2 \times 2 \times 3= 2^2 \times 3$, we can rewrite it as $\sqrt{12} = \sqrt{ 2^2 \times 3 } = \sqrt{2^2} \times \sqrt{3} = 2 \sqrt{3}.$ $_\square$

Simplify $\sqrt{72}$.

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See Solution

First, ask yourself, "What is a perfect square factor of $72$?"

$4$ is a perfect square factor of $72$, and $9$ is a perfect square factor of $72$.

For the sake of this process, it is more efficient to find the largest perfect square factor of $72$. As shown below, $4\times 9=36$ is the largest perfect square factor of $72:$

$$\begin{aligned} \sqrt{72}&=\sqrt{36\times 2} \\ &=\sqrt{36}\times\sqrt{2} \\ &=6\sqrt{2}. \end{aligned}$$

Therefore, simplified form of $\sqrt{72}$ is $6\sqrt{2}.\ _\square$

Note: When a number is placed to the left of a square root symbol, multiplication is implied. $“\, 6\sqrt{2}\, ”$ is read as $“\, 6$ times the square root of $2.\, ”$

Similarly, roots of higher degree (cube roots, fourth roots, etc.) are simplified when they have no factors under the radical that are perfect powers of the same degree as the radical.

Simplify $\sqrt[3] { a^2 b^ 4 }$.

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Notice that we have $b^3$, which is a cube factor in the radicand. Hence, we can pull it out to obtain

$$\sqrt[3] { a^2 b^4 } = b \sqrt[3] { a^2b}.\ _\square$$

Since radicals are actually exponential expressions, they follow the rules of exponents and cannot be added together. In particular, you should avoid the common mistake shown below:

$$\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}.$$

This means that when we are dealing with radicals with different radicands, like $\sqrt{5}$ and $\sqrt{7}$, there is really no way to combine or simplify them. However, when dealing with radicals that share a base, we can simplify them by combining like terms.

Simplify $\sqrt{12} + 3\sqrt{3}$.

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Since $\sqrt{12} = 2\sqrt{3}$, we have $\sqrt{12} + 3\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}.$ $_\square$

When multiplying radicals, we make extensive use of the identity $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$. This means that two radicals, when multiplied together, might produce an integer rather than another radical.

Simplify $\sqrt{12} \times \sqrt{3}$.

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Since $12 \times 3 = 36$, we have $\sqrt{12} \times \sqrt{3} = \sqrt{36} = 6$. $_\square$

This will not always happen; however, the identity used above can be applied even to more complicated radical multiplication.

Expand $\big(1+\sqrt{21}\big)\big( 3-\sqrt{3}\big)$.

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See Solution

Using the distributive property and the identity $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$, we have

$$\begin{aligned} \big(1+\sqrt{21}\big)\big( 3-\sqrt{3}\big) &= 1\big(3-\sqrt{3}\big) + \sqrt{21}\big(3-\sqrt{3}\big) \\ &= 3-\sqrt{3} + 3\sqrt{21} - \sqrt{21} \sqrt{3} \\ &= 3 - \sqrt{3} + 3\sqrt{21} - \sqrt{63} \\ &= 3 - \sqrt{3} + 3\sqrt{21} - 3\sqrt{7}.\ _\square \end{aligned}$$

For more detail, refer to Rationalizing Denominators.

Fractions are not considered to be written in simplest form if they have an irrational number $\big($like $\sqrt{2}$, for example$\big)$ in the denominator. We can simplify the fraction by rationalizing the denominator. This is a procedure that frequently appears in problems involving radicals.

For problems involving simple radicals, the approach is fairly simple.

Simplify $\frac{2}{\sqrt{3}}$.

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We can simplify this fraction by multiplying by $1=\frac{\sqrt{3}}{\sqrt{3}}$. Thus, we have $\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}= \frac{2\sqrt{3}}{3}$. $_\square$

When we have a more complicated expression involving a radical, we must look for another approach. However, when the denominator is a binomial expression involving radicals, we can use the difference of two squares identity to produce a conjugate pair that will remove the radicals from the denominator. For example, if we want to remove the radicals from the expression $\sqrt{2}+\sqrt{3}$, we can multiply it by its conjugate pair $\sqrt{2} - \sqrt{3}$.

Simplify $\frac{1}{\sqrt{2}+\sqrt{3}}$.

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We have

$$\frac{1}{\sqrt{2}+\sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{\big(\sqrt{2}\big)^2 - \sqrt{6} + \sqrt{6} - \big(\sqrt{3}\big)^2} = \frac{\sqrt{2} - \sqrt{3}}{2-3} = \frac{\sqrt3-\sqrt2}{1} = \sqrt3 - \sqrt2.\ _\square$$

• Simplifying Expressions

• Radical Equations

• Rationalizing Denominators