Question

Find $a$ and $b$ if $\dfrac{{4 - 2\sqrt 3 }}{{2 + \sqrt 3 }} = a + b\sqrt 3 $

Answer 1

Hint: We will rationalise the given expression by multiplying the numerator and denominator by $2 - \sqrt 3 $. Then, simplify the expression and apply identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ to simplify denominator. Then, compare the resultant expression with $a + b\sqrt 3 $ to find the value of $a$ and $b$.

Complete step-by-step answer:
We begin by rationalising the denominator of the given expression, $\dfrac{{4 - 2\sqrt 3 }}{{2 + \sqrt 3 }}$
We will multiply and divide by $2 - \sqrt 3 $ of the given expression.
$\dfrac{{4 - 2\sqrt 3 }}{{2 + \sqrt 3 }} \times \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}$
On simplifying the given expression using the identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator, we will get,
$\dfrac{{8 - 4\sqrt 3 - 4\sqrt 3 + 6}}{{{{\left( 2 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} = \dfrac{{14 - 8\sqrt 3 }}{{4 - 3}} = \dfrac{{14 - 8\sqrt 3 }}{1}$
Therefore, $\dfrac{{4 - 2\sqrt 3 }}{{2 + \sqrt 3 }}$ is equivalent to $14 - 8\sqrt 3 $.
Also, we are given $\dfrac{{4 - 2\sqrt 3 }}{{2 + \sqrt 3 }} = a + b\sqrt 3 $
This implies
$14 - 8\sqrt 3 = a + b\sqrt 3 $
On comparing the above equation, we get $a = 14,b = - 8$.

Note: Rationalisation is a process by which we convert any irrational number into a rational number. We multiply and divide such that we can use the identity $\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$ to convert the number into rational number.