Multiplication of surds | How to multiply surds

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.

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

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

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

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

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



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 $\sqrt{7}$


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

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

$=2 \times \sqrt{21}$



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


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

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

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



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

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


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.

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

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

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

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

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


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



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