Question

Simplify the expression: $ \dfrac{{4 + \sqrt 5 }}{{4 - \sqrt 5 }} + \dfrac{{4 - \sqrt 5 }}{{4 + \sqrt 5 }} $
A. $ \dfrac{{42}}{{11}} $
B. $ \dfrac{{40}}{{11}} $
C. $ \dfrac{{39}}{{25}} $
D. $ \dfrac{{16}}{{25}} $

Answer 1

Hint: There are two types of numbers; real numbers are further divided into two parts – Rational and Irrational numbers. The numbers which cannot be expressed as a ratio of two integers are called irrational numbers. As there is an irrational number in the denominator of the given numbers so we rationalize them and then add the two numbers.

Complete step-by-step answer:
First we rational both the terms that are in addition with each other –
Rationalising $ \dfrac{{4 + \sqrt 5 }}{{4 - \sqrt 5 }} $ , we get –
 $\Rightarrow \dfrac{{4 + \sqrt 5 }}{{4 - \sqrt 5 }} \times \dfrac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }} \\
 = \dfrac{{{{(4 + \sqrt 5 )}^2}}}{{{{(4)}^2} - {{(\sqrt 5 )}^2}}} \\
 = \dfrac{{{{(4 + \sqrt 5 )}^2}}}{{16 - 5}} = \dfrac{{{{(4 + \sqrt 5 )}^2}}}{{11}} $
Rationalising $ \dfrac{{4 - \sqrt 5 }}{{4 + \sqrt 5 }} $ , we get -
 $ \dfrac{{4 - \sqrt 5 }}{{4 + \sqrt 5 }} \times \dfrac{{4 - \sqrt 5 }}{{4 - \sqrt 5 }} \\
= \dfrac{{{{(4 - \sqrt 5 )}^2}}}{{{{(4)}^2} - {{(\sqrt 5 )}^2}}}\\
= \dfrac{{{{(4 - \sqrt 5 )}^2}}}{{16 - 5}} = \dfrac{{(4 - \sqrt 5 )}^2}{{11}} $
Now add the above two rationalized terms –
$\Rightarrow \dfrac{{{{(4 + \sqrt 5 )}^2}}}{{11}} + \dfrac{{{{(4 - \sqrt 5 )}^2}}}{{11}} \\
= \dfrac{{{{(4)}^2} + (\sqrt 5 ) + 2 \times 4 \times \sqrt 5 + {{(4)}^2} + {{(\sqrt 5 )}^2} - 2 \times 4 \times \sqrt 5 }}{{11}} \\
= \dfrac{{16 + 5 + 8\sqrt 5 + 16 + 5 - 8\sqrt 5 }}{{11}} = \dfrac{{42}}{{11}} $
Thus, $ \dfrac{{4 + \sqrt 5 }}{{4 - \sqrt 5 }} + \dfrac{{4 - \sqrt 5 }}{{4 + \sqrt 5 }} = \dfrac{{42}}{{11}} $
So, the correct answer is “Option A”.

Note: When we have a fraction in which the denominator is an irrational number, we rationalize it by multiplying both the numerator and denominator with the denominator but with opposite signs. We do rationalization to make the calculation simpler as the irrational term is converted to rational. We can also solve the question by simply taking LCM in the denominator as follows -
 $ \dfrac{{4 + \sqrt 5 }}{{4 - \sqrt 5 }} + \dfrac{{4 - \sqrt 5 }}{{4 + \sqrt 5 }} = \dfrac{{(4 + \sqrt 5 )(4 - \sqrt 5 ) + (4 - \sqrt 5 )(4 - \sqrt 5 )}}{{(4 - \sqrt 5 )(4 + \sqrt 5 )}} = \dfrac{{{{(4 + \sqrt 5 )}^2} + {{(4 - \sqrt 5 )}^2}}}{{{{(4)}^2} - {{(\sqrt 5 )}^2}}} = \dfrac{{42}}{{11}} $