Question

Simplify each of the following by rationalizing the denominator:
(i). $\dfrac{{6 - 4\sqrt 2 }}{{6 + 4\sqrt 2 }}$
(ii). $\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }}$
(iii). $\dfrac{1}{{3\sqrt 2 - 2\sqrt 3 }}$

Answer 1

Hint: We start solving this problem by equating the term to a variable. Then multiplying the numerator and denominator of the given term with the conjugate of the denominator by making use of the fact that conjugate of the irrational number $a + \sqrt b $ as $a - \sqrt b $. We use the fact that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ to proceed through the problem.

Complete step-by-step solution:
(i). $\dfrac{{6 - 4\sqrt 2 }}{{6 + 4\sqrt 2 }}$
By Taking conjugate,
$ \dfrac{{6 - 4\sqrt 2 }}{{6 + 4\sqrt 2 }} = \dfrac{{6 - 4\sqrt 2 }}{{6 + 4\sqrt 2 }} \times \dfrac{{6 - 4\sqrt 2 }}{{6 - 4\sqrt 2 }}$
 $ = \dfrac{{{{\left( {6 - 4\sqrt 2 } \right)}^2}}}{{{{(6)}^2} - {{(4\sqrt 2 )}^2}}} \\
   = \dfrac{{36 + 32 - 48\sqrt 2 }}{4} \\
   = 17 - 12\sqrt 2 $ (By using the formulae ${(a - b)^2} = {a^2} + {b^2} - 2ab$)
Therefore, The simplified form of the given term $\dfrac{{6 - 4\sqrt 2 }}{{6 + 4\sqrt 2 }}$ is $17 - 12\sqrt 2 $
(ii). $\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }}$
By Taking conjugate,
$ \dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} = \dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} \times \dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} $
$= \dfrac{{{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}} \\
   = \dfrac{{7 + 5 - 2\sqrt {35} }}{2} \\
   = 6 - \sqrt {15} \\
 $(By using the formulae ${\left( {a + b} \right)^2} = {a^2} + {b^2} - 2ab$)
Therefore, the simplified form of the given term $\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }}$ is $6 - \sqrt {15} $
(iii). $\dfrac{1}{{3\sqrt 2 - 2\sqrt 3 }}$
By taking conjugate we get,
$\dfrac{1}{{3\sqrt 2 - 2\sqrt 3 }} = \dfrac{1}{{3\sqrt 2 - 2\sqrt 3 }} \times \dfrac{{3\sqrt 2 + 2\sqrt 3 }}{{3\sqrt 2 + 2\sqrt 3 }}$
$= \dfrac{{3\sqrt 2 + 2\sqrt 3 }}{{{{\left( {3\sqrt 2 } \right)}^2} - {{\left( {2\sqrt 3 } \right)}^2}}} \\
   = \dfrac{{3\sqrt 2 + 2\sqrt 3 }}{6} $
Therefore, the simplified form of the given term $\dfrac{1}{{3\sqrt 2 - 2\sqrt 3 }}$ is $\dfrac{{3\sqrt 2 + 2\sqrt 3 }}{6}$

Note: Here are the six ways to rationalize the denominator:
Step1: For rationalizing the denominator, you have to multiply both numerator and the denominator by the conjugation of the denominator. To find the conjugate you have to change the sign of the denominator (between the two terms).
Step2: here distribute both the numerator and the denominator. Now you can multiply the numbers outside the radical with numbers outside the radical and also multiply numbers inside the radical with numbers inside the radical.
Step3: Combine the both terms
Step4: Simplifying the radicals
Step5: Again combining the terms
Step6: Now you can reduce the fraction. To reduce the fraction part, you have to reduce each number outside the radical by the same number. If you can't reduce the fraction, that is your final answer.