Question

Solve the following expression and convert it into most reducible form:
$\sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{4}}}$

Answer 1

Hint: The number given in the question is an irrational number. Simplify it and convert it in the form of $\dfrac{p}{q}$. Then rationalise the dfraction to get the most reducible form.

Complete step by step answer:
From the question, we have to determine the value of $\sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{4}}}$.
Let its value is $x$. Then we have:
$ \Rightarrow x = \sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{4}}} .....(i)$
Simplifying it further, we’ll get:
\[
   \Rightarrow x = \sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{\dfrac{{4 + \sqrt 3 }}{4}}} \\
   \Rightarrow x = \sqrt 3 + \dfrac{{\dfrac{1}{1}}}{{\dfrac{{4 + \sqrt 3 }}{2}}} \\
   \Rightarrow x = \sqrt 3 + \dfrac{2}{{4 + \sqrt 3 }} \\
\]
Cross multiplying \[\sqrt 3 \] with \[4 + \sqrt 3 \], we’ll get
\[
   \Rightarrow x = \dfrac{{4\sqrt 3 + 3 + 2}}{{4 + \sqrt 3 }} \\
   \Rightarrow x = \dfrac{{5 + 4\sqrt 3 }}{{4 + \sqrt 3 }} \\
\]
On rationalising the above expression, we will get:
\[ \Rightarrow x = \dfrac{{5 + 4\sqrt 3 }}{{4 + \sqrt 3 }} \times \dfrac{{4 - \sqrt 3 }}{{4 - \sqrt 3 }},\]
In the denominator, we are getting the form $\left( {a + b} \right)\left( {a - b} \right)$.
So using $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ and simultaneously multiplying numbers in numerator step by step, we’ll get
\[
   \Rightarrow x = \dfrac{{20 - 5\sqrt 3 + 16\sqrt 3 - 4\sqrt 3 \times \sqrt 3 }}{{{{\left( 4 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}, \\
   \Rightarrow x = \dfrac{{20 - 12 + 11\sqrt 3 }}{{16 - 3}}, \\
   \Rightarrow x = \dfrac{{8 + 11\sqrt 3 }}{{13}} \\
\]
Substituting the value of $x$from equation $(i)$, we have:
$\sqrt 3 + \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{{\sqrt 3 }}{4}}} = \dfrac{{8 + 11\sqrt 3 }}{{13}}$

Hence, the value of the above expression is \[\dfrac{{8 + 11\sqrt 3 }}{{13}}\].

Note: If in any expression, we are getting an irrational number in the denominator, we can always rationalise the number by multiplying and dividing with the conjugate of the denominator. In that case, the denominator will become a rational number.