In this section, we will learn how to multiply two or more surds. This is a topic on multiplication of surds grade 11, 12. For the basics of surds, please visit our page an introduction to surds.

Table of Contents

**How to Multiply Surds**

The multiplication of surds is one of the basic operations on surds. We need to follow the below steps while multiplying surds.

Step I: At first, we need to find the simplest forms of the surds involved in the multiplication.

Step II: Next, we check whether their simplest forms are of the same order or not.

Step III: In this step, we will separate the same order surds and the different order surds.

Step IV: Now we will multiply the same order surds as follows. Let us multiply their rational coefficients and then multiply the surd-factors, and keep them side by side.

Step V: To multiply the surds of different orders, we need to reduce them into surds of the same order, and then multiply them by the above rules.

Step VI: For the multiplication of different order surds with the same base, see the remark in the end.

**Example of how to multiply surds:** Let us now understand the above process with an example. Find the value of $3\sqrt{2}$ $\times \sqrt{8}.$ Look that $\sqrt{2}$ is already in the simplest form as we cannot further simplify it. Now simplify $\sqrt{8}.$ Note that $8=$ $2 \times 2 \times 2.$ So we have

$\sqrt{8}=\sqrt{2 \times 2} \times \sqrt{2}$ $=2\sqrt{2}.$

Observe that the simplest forms of both $\sqrt{2}$ and $\sqrt{8}$ have order $2,$ that is, they are of the same order. So according to the above Step IV, we get that

$3\sqrt{2} \times \sqrt{8}$

$=3\sqrt{2} \times 2\sqrt{2}$

$=(3 \times 2) \times \sqrt{2 \times 2}$

$=6 \times 2$ $[\because \sqrt{a \times a}=a]$

$=12$

**Formulas of Multiplication of Surds**

From the above method of finding the product of two or more surds, the following rules can be derived for the multiplication of surds.

(i) $\sqrt{a} \times \sqrt{b}=\sqrt{ab}$

(ii) $\sqrt[n]{a} \times \sqrt[n]{b}=\sqrt[n]{ab}$

(iii) $x\sqrt{a} \times y\sqrt{b}=xy\sqrt{ab}$

(iv) $x\sqrt[n]{a} \times y\sqrt[n]{b}=xy\sqrt[n]{ab}$

(iv) $\sqrt[m]{a} \times \sqrt[n]{a}=a^{\frac{1}{m}+\frac{1}{n}}$

(v) $x\sqrt[m]{a} \times y\sqrt[n]{a}=xy \times a^{\frac{1}{m}+\frac{1}{n}}$

**Solved Problems of Multiplication of Surds**

**Problem 1:** Lets multiply $2\sqrt{3}$ and $\sqrt{7}$

Solution:

$2\sqrt{3} \times \sqrt{7}$

$=(2 \times 1) \times \sqrt{3 \times 7}$

$=2 \times \sqrt{21}$

$=2\sqrt{21}$

**Problem 2:** Multiply $5\sqrt[3]{2}$ and $2\sqrt[3]{3}$

Solution:

$5\sqrt[3]{2} \times 2\sqrt[3]{3}$

$=(5 \times 2) \times \sqrt[3]{2 \times 3}$

$=10 \times \sqrt[3]{6}$

$=10\sqrt{12}$

**Remark:** If the surds involved in the multiplication have the same base but different orders, then we multiply them according to the rules of indices. See the example below.

**Example: **Simplify the expression

$\sqrt{2} \times 3 \sqrt[4]{8} \times 2\sqrt[3]{16}$

Solution:

Note that the surds $\sqrt{2},$ $\sqrt[4]{8}$ and $\sqrt[3]{16}$ have different orders, but we can check that they have the same after simplifying the surds.

$\sqrt[4]{8}$ $=8^{1/4}$ $=(2^3)^{1/4}$ $=2^{3/4}$

$\sqrt[3]{16}$ $=16^{1/3}$ $=(2^4)^{1/3}$ $=2^{4/3}$

∴ $\sqrt{2} \times 3 \sqrt[4]{8} \times 2\sqrt[3]{16}$

$=2^{1/2} \times 2^{3/4} \times 2^{4/3}$

$=2^{\frac{1}{2}+\frac{3}{4}+\frac{4}{3}}$ $[\because a^m.a^n.a^p=a^{m+n+p}]$

$=2^\frac{31}{12}$

$=\sqrt[12]{2^{31}}$ $[\because a^{\frac{m}{n}}=\sqrt[n]{a^m}]$

$=\sqrt[12]{2147483648}$

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