Question

How do you simplify $\sqrt{\left( \dfrac{1}{18} \right)}$ ?

Answer 1

Hint: We are asked to simplify $\sqrt{\left( \dfrac{1}{18} \right)}$, we will first learn what things need to be simplified. We learn about cancelling common term, then we go for cancelling radical in the denominator as our term $\sqrt{\left( \dfrac{1}{18} \right)}$ has $\sqrt{18}$ radical in denominator so we use that cancellation process to simplify in this we multiply and divide term by $\sqrt{3}$ and then we get after multiplication will be the required term.

Complete answer:
We are given $\sqrt{\left( \dfrac{1}{18} \right)}$. We are asked to simplify and we have to take the step to simplify.
Before we solve it we must learn what we need to solve such a problem.
If we are given problem in a fraction like $\dfrac{a}{b}$ , we must check that numerator and denominator ‘b’ has something common or not, if they have some common term for example consider $\dfrac{18}{4}$ , we have numerator as ‘18’ and denominator as ‘4’.
We can see that the numerator and denominator have ‘2’ as common because $18=2\times 9$ and $4=2\times 2$ , so we first cancel that 2.
So, we get –
$\dfrac{18}{4}=\dfrac{2\times 9}{2\times 2}=\dfrac{9}{2}$
So we get –
$\dfrac{18}{4}$ simplify and become $\dfrac{9}{2}$ .
Our term is given as $\sqrt{\left( \dfrac{1}{18} \right)}$, we simplify it first we get that $\sqrt{1}=1$ so we get –
$\sqrt{\dfrac{1}{18}}=\dfrac{1}{\sqrt{18}}$
We factor the denominator,
We have ‘18’ in the denominator and it is factorized as $2\times 3\times 3$ . So we get –
$\sqrt{18}=\sqrt{2\times 3\times 3}$ , ‘3’ is making a pair so 3 will come out of radical.
Hence, we get –
$\sqrt{18}=3\sqrt{2}$
Therefore, $\sqrt{\left( \dfrac{1}{18} \right)}=\dfrac{1}{3\sqrt{2}}$ .
Second thing to check is that, denominator must have radical expression like $\sqrt{b}$ ,
If we have number like $\sqrt{b}$ , in denominator then we multiply and divide the term by $\sqrt{b}$ , and simplify like if we have term as $\dfrac{a}{\sqrt{b}}$ , so we multiply by $\sqrt{b}$ and divide by $\sqrt{b}$ , we get –
$\dfrac{a}{\sqrt{b}}=\dfrac{a\times \sqrt{b}}{\sqrt{b}\times \sqrt{b}}=\dfrac{a\sqrt{b}}{b}$
Now we go to cancel radical present in denominator because radical cannot be these in denominator in simplest form.
So, we have $\sqrt{2}$ in denominator, we multiply and divide by $\sqrt{2}$ so we get –
$\sqrt{\dfrac{1}{18}}=\dfrac{1}{3\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$
As $\sqrt{2}\times \sqrt{2}=2$ so we get –
$=\dfrac{1\times \sqrt{2}}{3\times 2}$
As $1\times \sqrt{2}=\sqrt{2}$ so we get –
$=\dfrac{\sqrt{2}}{6}$
So simplest form of $\sqrt{\dfrac{1}{18}}$ is $\dfrac{\sqrt{2}}{6}$

Note: If we have two term in denominator with radical like $\dfrac{a}{b+\sqrt{c}}$ , then we will multiply and divide by $b-\sqrt{c}$ , so we will have –
$\dfrac{a}{b+\sqrt{c}}=\dfrac{a}{b+\sqrt{c}}\times \dfrac{\left( b-\sqrt{c} \right)}{\left( b-\sqrt{c} \right)}$
By simplifying, we get –
$=\dfrac{ab-a\sqrt{c}}{{{b}^{2}}-c}$ , this will be our simplified form.