Question

Find the value of a & b $\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} = a + \dfrac{{7\sqrt 5 }}{{11}}b$.

Answer 1

Hint: First we will rationalize the denominator of L.H.S of rational number. Further, we will simplify and write in the same way as in value of the RHS. Thereafter we will compare the coefficient of $aandb$.


Complete step by step Solution:
 Given $\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }}\, - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} = a + \dfrac{{7\sqrt 5 }}{{11}}b$
We will take L.H.S.,we have
$\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }}$
Rationalizing the denominator of both the terms , we get
$\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} \times \dfrac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} \times \dfrac{{7 - \sqrt 5 }}{{7 - \sqrt 5 }}$
$ = \dfrac{{{{\left( {7 + \sqrt 5 } \right)}^2}}}{{{7^2} - {{\left( {\sqrt 5 } \right)}^2}}} - \dfrac{{{{\left( {7 - \sqrt 5 } \right)}^2}}}{{{7^2} - {{\left( {\sqrt 5 } \right)}^2}}}$ …(i)
 By using the algebraic identity used in (i) equation
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab,\,\,{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ and ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ .
So $\dfrac{{{7^2} + {{\left( {\sqrt 5 } \right)}^2} + 2 \times 7 \times \sqrt 5 }}{{49 - 5}} - \dfrac{{{7^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times 7 \times \sqrt 5 }}{{49 - 5}}$
$ = \dfrac{{49 + 5 + 14\sqrt 5 }}{{44}} - \dfrac{{49 + 5 - 14\sqrt 5 }}{{44}}$
$\dfrac{{54 + 14\sqrt 5 }}{{44}} - \dfrac{{54 - 14\sqrt 5 }}{{44}}$
$ = \dfrac{{54 + 14\sqrt 5 - 54 + 14\sqrt 5 }}{{44}}$
$ = \dfrac{{8\sqrt 5 }}{{44}} = \dfrac{{7\sqrt 5 }}{4}$
$ = 0 + \dfrac{{7\sqrt 5 }}{4}$
Hence $0 + \dfrac{{7\sqrt 5 }}{4} = $ R.H.S.
$0 + \dfrac{{7\sqrt 5 }}{4} = a + \dfrac{{7\sqrt 5 }}{4}b$
Comparing both sides we get $a = 0\,and\,\,b = 1.$


Note: In this type of question,students must keep in mind the uses of algebraic identities and use suitable identity for any value. When we rationalize the denominator, we take the opposite sign of the value in the denominator in the numerator.