Question

Find the value of
$\sqrt {\dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}} $

Answer 1

Hint: To solve the given question, we will take $\sqrt {\dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}} $ is equal to x. Then by squaring both sides, we can remove the square root which will help in solving the question easily. After that removing the square root from the denominator by rationalization method. In the rationalisation method, we will multiply the numerator and denominator by the number i.e. $33 + 19\sqrt 3 $.

Complete step-by-step answer:
First of all, we take the value of $\sqrt {\dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}} $ as $x$ i.e.
$ \Rightarrow x = \sqrt {\dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}} $
By squaring both sides, we get,
$ \Rightarrow {x^2} = {\left( {\sqrt {\dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}} } \right)^2}$
By cancelling square root with square, we get,
$ \Rightarrow {x^2} = \dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }}$
Now, we will do the rationalisation to eliminate the square root from the denominator. To eliminate square root from denominator we will multiply the numerator and denominator by $33 + 19\sqrt 3 $ and hence, we get,
$ \Rightarrow {x^2} = \dfrac{{6 + 2\sqrt 3 }}{{33 - 19\sqrt 3 }} \times \dfrac{{33 + 19\sqrt 3 }}{{33 + 19\sqrt 3 }}$
$ \Rightarrow {x^2} = \dfrac{{\left( {6 + 2\sqrt 3 } \right) \times \left( {33 + 19\sqrt 3 } \right)}}{{\left( {33 - 19\sqrt 3 } \right) \times \left( {33 + 19\sqrt 3 } \right)}}$
In denominator the equation is in form of $\left( {a - b} \right)\left( {a + b} \right)$ and $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$. Hence, we can write above equation as:
$ \Rightarrow {x^2} = \dfrac{{\left( {6 + 2\sqrt 3 } \right) \times \left( {33 + 19\sqrt 3 } \right)}}{{{{33}^2} - {{\left( {19\sqrt 3 } \right)}^2}}}$
In numerator open the bracket and apply distributive property, we get,
$ \Rightarrow {x^2} = \dfrac{{33\left( {6 + 2\sqrt 3 } \right) + 19\sqrt 3 \left( {6 + 2\sqrt 3 } \right)}}{{{{33}^2} - {{\left( {19\sqrt 3 } \right)}^2}}}$
Now, we will solve the numerator by opening the bracket and multiply the number with both as follows:
$ \Rightarrow {x^2} = \dfrac{{33 \times 6 + 33 \times 2\sqrt 3 + 19\sqrt 3 \times 6 + 19\sqrt 3 \times 2\sqrt 3 }}{{{{33}^2} - {{\left( {19\sqrt 3 } \right)}^2}}}$
$ \Rightarrow {x^2} = \dfrac{{198 + 66\sqrt 3 + 114\sqrt 3 + 38 \times 3}}{{{{33}^2} - {{\left( {19\sqrt 3 } \right)}^2}}}$
$ \Rightarrow {x^2} = \dfrac{{198 + 66\sqrt 3 + 114\sqrt 3 + 114}}{{{{33}^2} - {{\left( {19\sqrt 3 } \right)}^2}}}$
Solve the denominator by squaring the numbers as the square of 33 is 1089 and square of $19\sqrt 3 $ is 1083. Hence, we get,
$ \Rightarrow {x^2} = \dfrac{{198 + 66\sqrt 3 + 114\sqrt 3 + 114}}{{1089 - 1083}}$
By adding the root 3 terms separately and others separately in numerator we get,
$ \Rightarrow {x^2} = \dfrac{{312 + 180\sqrt 3 }}{{1089 - 1083}}$
By subtracting the numbers in denominator, we get,
$ \Rightarrow {x^2} = \dfrac{{312 + 180\sqrt 3 }}{6}$
By taking 6 common from denominator we get,
$ \Rightarrow {x^2} = \dfrac{{6 \times 52 + 6 \times 30\sqrt 3 }}{6}$
$ \Rightarrow {x^2} = \dfrac{{6 \times \left( {52 + 30\sqrt 3 } \right)}}{6}$
By cancelling 6 from numerator with denominator we get,
$ \Rightarrow {x^2} = 52 + 30\sqrt 3 $
We can split the numerator as given below,
$ \Rightarrow {x^2} = 52 + 30\sqrt 3 $
$ \Rightarrow {x^2} = 25 + 27 + 30\sqrt 3 $
We can write 25 as squares of 5 and 27 as squares of $3\sqrt 3 $. Also, $30\sqrt 3 $ can be written as $2 \times 5 \times 3\sqrt 3 $ hence we get,
$ \Rightarrow {x^2} = {5^2} + {\left( {3\sqrt 3 } \right)^2} + 2 \times 5 \times 3\sqrt 3 $
$ \Rightarrow {x^2} = {\left( {5 + 3\sqrt 3 } \right)^2}$
By taking square root on both sides
$ \Rightarrow x = 5 + 3\sqrt 3 $
Hence, the required answer is $5 + 3\sqrt 3 $.

Note: Students can make mistakes while rationalizing the method, they can multiply the numerator and denominator by different numeric or they can forget to multiply with the numerator and only do multiplication of denominator. By doing this their answer will get wrong.
Secondly, some students multiply them with the numeric in denominator without changing the sign but if we want to remove square root from denominator then sign should be of only one term out of two terms only then we can apply the formula of $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$.