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Question: If \[a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\] and \[b = \dfrac{{\sqrt 5 - \sqrt 2 }}...

If a=5+25−2a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} and b=5−25+2b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }} , find the value of a2+ab+b2a2−ab+b2\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}

Explanation

Solution

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), (a−b)(a + b),{\text{ }}(a - b) and abab. Now write a2+ab+b2a2−ab+b2\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}as (a+b)2−ab(a−b)2+ab\dfrac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a - b} \right)}^2} + ab}} and substitute the values of (a+b), (a−b)(a + b),{\text{ }}(a - b) and abab, on solving we get our answer.

Complete step by step answer:

Given, a=5+25−2a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} and b=5−25+2b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}
Therefore, ab=5+25−2×5−25+2=1ab = \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=5+25−2a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}
Multiplying both the numerator and denominator with 5+2\sqrt 5 + \sqrt 2 , we get
⇒a=5+25−2×5+25+2\Rightarrow a = \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }} \times \dfrac{{\sqrt 5 + \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}
⇒a=(5+2)2(5−2)(5+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)=a2−b2(a + b)(a - b) = {a^2} - {b^2}, we get
⇒a=(5)2+(2)2+2×5×2(5)2−(2)2\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,
⇒a=5+2+2105−2\Rightarrow a = \dfrac{{5 + 2 + 2\sqrt {10} }}{{5 - 2}}
On addition we get,
⇒a=7+2103\Rightarrow a = \dfrac{{7 + 2\sqrt {10} }}{3}
And b=5−25+2b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }}
Multiplying both the numerator and denominator with 5−2\sqrt 5 - \sqrt 2 ,
⇒b=5−25+2×5−25−2\Rightarrow b = \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 + \sqrt 2 }} \times \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}
⇒b=(5−2)2(5−2)(5+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)=a2−b2(a + b)(a - b) = {a^2} - {b^2}, we get
⇒b=(5)2+(2)2−2×5×2(5)2−(2)2\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,
⇒b=5+2−2105−2\Rightarrow b = \dfrac{{5 + 2 - 2\sqrt {10} }}{{5 - 2}}
On addition we get,
⇒b=7−2103\Rightarrow b = \dfrac{{7 - 2\sqrt {10} }}{3}
Now, calculating a+ba + b,
We get,
a+b=7+2103+7−2103a + b = \dfrac{{7 + 2\sqrt {10} }}{3} + \dfrac{{7 - 2\sqrt {10} }}{3}
On simplification we get,
⇒a+b=7+210+7−2103\Rightarrow a + b = \dfrac{{7 + 2\sqrt {10} + 7 - 2\sqrt {10} }}{3}
⇒a+b=143\Rightarrow a + b = \dfrac{{14}}{3}
And,
On calculating a−ba - b,
We get,
a−b=7+2103−7−2103a - b = \dfrac{{7 + 2\sqrt {10} }}{3} - \dfrac{{7 - 2\sqrt {10} }}{3}
On simplification we get,
⇒a−b=7+210−7+2103\Rightarrow a - b = \dfrac{{7 + 2\sqrt {10} - 7 + 2\sqrt {10} }}{3}
⇒a−b=4103\Rightarrow a - b = \dfrac{{4\sqrt {10} }}{3}
Now we have to find the value of a2+ab+b2a2−ab+b2\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}
Therefore,
a2+ab+b2a2−ab+b2\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}
Add and subtract abab, from numerator and denominator, we get
=a2+2ab+b2−aba2−2ab+b2+ab= \dfrac{{{a^2} + 2ab + {b^2} - ab}}{{{a^2} - 2ab + {b^2} + ab}}
On reframing we get,
=(a+b)2−ab(a−b)2+ab= \dfrac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a - b} \right)}^2} + ab}}
On Substituting the values of (a+b), (a−b)(a + b),{\text{ }}(a - b) and abab, we get
=(143)2−1(4103)2+1= \dfrac{{{{\left( {\dfrac{{14}}{3}} \right)}^2} - 1}}{{{{\left( {\dfrac{{4\sqrt {10} }}{3}} \right)}^2} + 1}}
On simplification we get,
=1969−11609+1= \dfrac{{\dfrac{{196}}{9} - 1}}{{\dfrac{{160}}{9} + 1}}
On taking LCM we get,
=196−9160+9= \dfrac{{196 - 9}}{{160 + 9}}
On simplification we get,
=187169= \dfrac{{187}}{{169}}
Therefore, the value of a2+ab+b2a2−ab+b2\dfrac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} is 187169\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 a+bc+d\dfrac{{a + \sqrt b }}{{c + \sqrt d }} , we will multiply both the numerator and the denominator with c−dc - \sqrt d .
⇒a+bc+d=a+bc+d×c−dc−d=(a+b)(c−d)c2−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)=a2−b2(a + b)(a - b) = {a^2} - {b^2} ]