Multiply Radicals - Questions with Solutions for Grade 10

Grade 10 questions on how to multiply expressions with radicals are presented below. Try to solve each expression, then click the arrow to view the step-by-step solutions.

Product (Multiplication) of Radicals With the Same Index

The multiplication formula for radicals with equal indices is given by:

$${\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{x \cdot y}}$$

Examples

Review the following examples to see how the product rule is applied:

  1. $\sqrt{3} \cdot \sqrt{12} = \sqrt{3 \cdot 12} = \sqrt{36} = 6$
  2. $\sqrt[3]{27x^3} = \sqrt[3]{27} \cdot \sqrt[3]{x^3} = 3x$
  3. $\sqrt[4]{\frac{1}{12}} \cdot \sqrt[4]{3} \cdot \sqrt[4]{64} = \sqrt[4]{\frac{1}{12} \cdot 3 \cdot 64} = \sqrt[4]{16} = 2$
  4. $3\sqrt[10]{x^3} \cdot 5\sqrt[10]{x^5} \cdot \sqrt[10]{x^2} = (3 \cdot 5) \sqrt[10]{x^3 \cdot x^5 \cdot x^2} = 15 \sqrt[10]{x^{10}} = 15|x| = 15x$ (assuming $x \geq 0$)

Practice Questions With Answers

Use the multiplication formula to simplify the following expressions:

Question 1

Simplify: $4\sqrt{2} \cdot 7\sqrt{32}$

View Solution

$$4\sqrt{2} \cdot 7\sqrt{32} = (4 \cdot 7)\sqrt{2 \cdot 32}$$

$$= 28\sqrt{64} = 28 \cdot 8 = 224$$

Question 2

Simplify: $6\sqrt{x} \cdot \frac{2}{3}\sqrt{x}$

View Solution

$$6\sqrt{x} \cdot \frac{2}{3}\sqrt{x} = \left(6 \cdot \frac{2}{3}\right)\sqrt{x \cdot x}$$

$$= 4\sqrt{x^2} = 4|x| = 4x \quad (\text{assuming } x \geq 0)$$

Question 3

Simplify: $2\sqrt[3]{\frac{1}{32}} \cdot \sqrt[3]{128} \cdot \sqrt[3]{16}$

View Solution

$$2\sqrt[3]{\frac{1}{32}} \cdot \sqrt[3]{128} \cdot \sqrt[3]{16} = 2\sqrt[3]{\frac{1}{32} \cdot 128 \cdot 16}$$

$$= 2\sqrt[3]{64} = 2 \cdot 4 = 8$$

Question 4

Simplify: $-8\sqrt[5]{x^2} \cdot 2\sqrt[5]{x^3}$

View Solution

$$-8\sqrt[5]{x^2} \cdot 2\sqrt[5]{x^3} = (-8 \cdot 2)\sqrt[5]{x^2 \cdot x^3}$$

$$= -16\sqrt[5]{x^5} = -16x$$

Question 5

Simplify: $\sqrt{(x-3)} \cdot \sqrt{(x-3)}$

View Solution

$$\sqrt{(x-3)} \cdot \sqrt{(x-3)} = \sqrt{(x-3)(x-3)} = \sqrt{(x-3)^2}$$

$$= |x-3| = x-3$$

(Note: Because $\sqrt{(x-3)}$ is a real number, $(x-3) \geq 0$, which allows us to simplify the absolute value $|x-3|$ directly to $x-3$.)

Question 6

Simplify: $\sqrt[8]{x} \cdot 5\sqrt[8]{x^4} \cdot 2\sqrt[8]{x^3}$

View Solution

$$\sqrt[8]{x} \cdot 5\sqrt[8]{x^4} \cdot 2\sqrt[8]{x^3} = (1 \cdot 5 \cdot 2)\sqrt[8]{x \cdot x^4 \cdot x^3}$$

$$= 10\sqrt[8]{x^8} = 10|x| = 10x \quad (\text{assuming } x \geq 0)$$

References and Links