Question

What’s \[3\sqrt {18} \] ?

Answer 1

Hint: In the given problem, we are asked to simplify \[3\] times the square root of \[18\] i.e., \[3\sqrt {18} \] . In order to solve this problem, we will first take the prime factors of \[18\] . Then we will substitute the factors in the given expression. After that we will simplify it using the formulas \[\sqrt {ab} = \sqrt a \sqrt b \] and \[\sqrt {{a^2}} = a\] .And hence, we will get the required result.

Complete step by step answer:
The given problem is to simplify \[3\sqrt {18} \] that means \[3\] times the square root of \[18\]. Now first of all we will take the prime factors of \[18\] i.e.,
\[ \Rightarrow 18 = 2 \times 3 \times 3\]
Therefore, we can write the given expression as
\[3\sqrt {18} = 3\sqrt {2 \times 3 \times 3} \]
\[ \Rightarrow 3\sqrt {18} = 3\sqrt {2 \times \left( {3 \times 3} \right)} \]
Now we know that \[\sqrt {a \times a} = \sqrt {{a^2}} \]
Here, \[3\] appears two times, so we will make it a square
Therefore, we get
\[ \Rightarrow 3\sqrt {18} = 3\sqrt {2 \times {3^2}} \]

Now we will apply the rule as
\[\sqrt {ab} = \sqrt a \sqrt b \]
Therefore, we can write
\[ \Rightarrow 3\sqrt {18} = 3\sqrt 2 \sqrt {{3^2}} \]
Now we will apply the rule as
\[\sqrt {{a^2}} = a\]
Therefore, we get
\[ \Rightarrow 3\sqrt {18} = 3\sqrt 2 \left( 3 \right)\]
On multiplying, we get
\[ \therefore 3\sqrt {18} = 9\sqrt 2 \]

Hence, we get \[3\sqrt {18} \] equals \[9\sqrt 2 \].

Note: Here the symbol \[\sqrt {} \] is called Radical and the number inside the symbol is called Radicand. Note that when we want to find the square root of some number, let it be \[a\] .If \[a\] is a perfect square then we can obtain the result directly. Otherwise, if \[a\] is not a perfect square, we factorise it and simplify it. Also note that the given number is an irrational number because \[3\sqrt {18} \] cannot be expressed in the form of \[\dfrac{p}{q}\] .And we can also say \[3\sqrt {18} = 9\sqrt 2 \] Here \[\sqrt 2 \] is irrational and \[9\] is rational and the product of a rational and an irrational number is never rational, hence it is an irrational number.