Question

If \[\sqrt 3 = 1.732\] and \[\sqrt 2 = 1.414{\text{ }}\], then find the value of \[\dfrac{1}{{\sqrt 3 - \sqrt 2 }}\]

Answer 1

Hint: Here we will first rationalize the given quantity and then use some identities and the given values to get the answer.
Rationalization of the expression means converting the surd denominator into an integer by multiplying the denominator as well as the numerator by the conjugate of the denominator.
The identity is given by:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step by step solution:
The given values are:
\[\sqrt 3 = 1.732\]
\[\sqrt 2 = 1.414{\text{ }}\]
We have to find the value of \[\dfrac{1}{{\sqrt 3 - \sqrt 2 }}\]
Let \[t = \dfrac{1}{{\sqrt 3 - \sqrt 2 }}\]
Hence we will first rationalize it in order to make denominator an integer therefore on rationalizing we get:
\[
  t = \dfrac{1}{{\sqrt 3 - \sqrt 2 }} \times \dfrac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \\
  t = \dfrac{{\sqrt 3 + \sqrt 2 }}{{\left( {\sqrt 3 - \sqrt 2 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}} \\
 \]
Now as we know that:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Using this identity for the above equation we get:
\[
  t = \dfrac{{\sqrt 3 + \sqrt 2 }}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \\
  t = \dfrac{{\sqrt 3 + \sqrt 2 }}{{3 - 2}} \\
  t = \dfrac{{\sqrt 3 + \sqrt 2 }}{1} \\
  t = \sqrt 3 + \sqrt 2 \\
 \]
Now substituting the given values we get:
\[
  t = \left( {1.732} \right) + \left( {1.414} \right) \\
  t = 3.146 \\
 \]

Hence the value of \[\dfrac{1}{{\sqrt 3 - \sqrt 2 }}\] is $3.146$.

Note:
This question can also be done by directly substituting the given values and then solving the fraction to get the required value.
\[\dfrac{1}{{\sqrt 3 - \sqrt 2 }}\]
Substituting the given values we get:
\[
   = \dfrac{1}{{1.732 - 1.414}} \\
   = \dfrac{1}{{0.318}} \\
   = 3.146 \\
 \]