Question

If \[a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\] and \[b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}\] , find the value of $\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$

Answer 1

Hint: First we will have to rationalise the denominators of both a and b. Note that a and b are conjugate surds. Therefore find \[(a + b),{\text{ }}(a - b)\] and \[ab\]. Now write $\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$as $\dfrac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a - b} \right)}^2} + ab}}$ and substitute the values of \[(a + b),{\text{ }}(a - b)\] and \[ab\], on solving we get our answer.

Complete step by step answer:

Given, \[a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\] and \[b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}\]
Therefore, \[ab = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} \times \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }} = 1\]
Now we will have to rationalize the denominators of both a and b first.
For this, we multiply the denominator with its conjugate surd.
\[a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\]
Multiplying both the numerator and denominator with \[\sqrt 5 + \sqrt 2 \], we get
\[ \Rightarrow a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} \times \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}\]
\[ \Rightarrow a = \dfrac{{{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}}\]
Using, $(a + b)(a - b) = {a^2} - {b^2}$, we get
\[ \Rightarrow a = \dfrac{{{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} + 2 \times \sqrt 5 \times \sqrt 2 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}\]
On simplification we get,
\[ \Rightarrow a = \dfrac{{5 + 2 + 2\sqrt {10} }}{{5 - 2}}\]
On addition we get,
\[ \Rightarrow a = \dfrac{{7 + 2\sqrt {10} }}{3}\]
And \[b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}\]
Multiplying both the numerator and denominator with \[\sqrt 5 - \sqrt 2 \],
\[ \Rightarrow b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }} \times \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\]
\[ \Rightarrow b = \dfrac{{{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}}\]
Using, $(a + b)(a - b) = {a^2} - {b^2}$, we get
\[ \Rightarrow b = \dfrac{{{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 2 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}\]
On simplification we get,
\[ \Rightarrow b = \dfrac{{5 + 2 - 2\sqrt {10} }}{{5 - 2}}\]
On addition we get,
\[ \Rightarrow b = \dfrac{{7 - 2\sqrt {10} }}{3}\]
Now, calculating \[a + b\],
We get,
$a + b = \dfrac{{7 + 2\sqrt {10} }}{3} + \dfrac{{7 - 2\sqrt {10} }}{3}$
On simplification we get,
$ \Rightarrow a + b = \dfrac{{7 + 2\sqrt {10} + 7 - 2\sqrt {10} }}{3}$
$ \Rightarrow a + b = \dfrac{{14}}{3}$
And,
On calculating \[a - b\],
We get,
$a - b = \dfrac{{7 + 2\sqrt {10} }}{3} - \dfrac{{7 - 2\sqrt {10} }}{3}$
On simplification we get,
$ \Rightarrow a - b = \dfrac{{7 + 2\sqrt {10} - 7 + 2\sqrt {10} }}{3}$
$ \Rightarrow a - b = \dfrac{{4\sqrt {10} }}{3}$
Now we have to find the value of $\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$
Therefore,
$\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$
Add and subtract \[ab\], from numerator and denominator, we get
$ = \dfrac{{{a^2} + 2ab + {b^2} - ab}}{{{a^2} - 2ab + {b^2} + ab}}$
On reframing we get,
$ = \dfrac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a - b} \right)}^2} + ab}}$
On Substituting the values of \[(a + b),{\text{ }}(a - b)\] and \[ab\], we get
$ = \dfrac{{{{\left( {\dfrac{{14}}{3}} \right)}^2} - 1}}{{{{\left( {\dfrac{{4\sqrt {10} }}{3}} \right)}^2} + 1}}$
On simplification we get,
$ = \dfrac{{\dfrac{{196}}{9} - 1}}{{\dfrac{{160}}{9} + 1}}$
On taking LCM we get,
$ = \dfrac{{196 - 9}}{{160 + 9}}$
On simplification we get,
$ = \dfrac{{187}}{{169}}$
Therefore, the value of $\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}$ is $\dfrac{{187}}{{169}}$.

Note: To rationalize the denominator of a fraction, we have to multiply both the numerator and the denominator with the conjugate surd of the denominator. For example, to rationalise the denominator of the fraction $\dfrac{{a + \sqrt b }}{{c + \sqrt d }}$ , we will multiply both the numerator and the denominator with $c - \sqrt d $ .
$ \Rightarrow \dfrac{{a + \sqrt b }}{{c + \sqrt d }} = \dfrac{{a + \sqrt b }}{{c + \sqrt d }} \times \dfrac{{c - \sqrt d }}{{c - \sqrt d }} = \dfrac{{\left( {a + \sqrt b } \right)\left( {c - \sqrt d } \right)}}{{{c^2} - d}}$ [since $(a + b)(a - b) = {a^2} - {b^2}$ ]