Question

$\sqrt{6}$ multiplied with $\sqrt{27}$ is equal to.

Answer 1

Hint: In this question we have been given two radical numbers for which we have to find the product. We will solve this question by first writing the numbers given inside the radical as a product of prime numbers. We will then use the property of radicals that $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$ and split the terms. We will then use the property that $\sqrt{a}\times \sqrt{a}=a$ to simplify the expression and get the required solution.

Complete step-by-step solution:
We have to find what $\sqrt{6}$ multiplied by $\sqrt{27}$ is equal to.
It can be written in the mathematical form as:
$\Rightarrow \sqrt{6}\times \sqrt{27}\to \left( 1 \right)$
We know that $6=3\times 2$ therefore, on substituting, we get:
$\Rightarrow \sqrt{6}=\sqrt{3\times 2}$
Now we know the property of radicals that $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$, therefore, we can write:
$\Rightarrow \sqrt{6}=\sqrt{3}\times \sqrt{2}\to \left( 2 \right)$
Now we know that $27$ can be written as $3\times 3\times 3$ therefore, on substituting, we get:
$\Rightarrow \sqrt{27}=\sqrt{3\times 3\times 3}$
On using the same property, we get:
$\Rightarrow \sqrt{27}=\sqrt{3}\times \sqrt{3}\times \sqrt{3}\to \left( 3 \right)$
On substituting equations $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 1 \right)$, we get:
$\Rightarrow \sqrt{3}\times \sqrt{2}\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}$
Now we know that $\sqrt{a}\times \sqrt{a}=a$ therefore, we get:
$\Rightarrow 3\times 3\times \sqrt{2}$
On multiplying the terms, we get:
$\Rightarrow 9\sqrt{2}$, which is the simplification.
Therefore, we can write $\sqrt{6}\times \sqrt{27}=9\sqrt{2}$, which is the required answer.

Note: In these types of questions the first thing to do is to solve the square roots and simplify them and then the operation of multiplication, addition, division or subtraction is to be performed. The various properties regarding radical expressions should be remembered. A radical can also be expressed in the form of an exponent where $\sqrt[n]{a}={{a}^{{}^{1}/{}_{n}}}$.