Question

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

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Hint: To solve the function which involve roots where factoring is not possible we have to use the process of rationalization and also use algebraic formula $\left( {a + b} \right){\text{ }}\left( {a - b} \right) = {a^2} - {b^2}$ whenever required.

Complete step by step solution:
Mark the equation and solve the equations separately by marking it as (I) and (II),
$\mathop {\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }}}\limits_{(I)} - \mathop {\dfrac{{7\sqrt 5 }}{{7 + \sqrt 5 }}}\limits_{(II)} = a + \dfrac{{7\sqrt 5 }}{{11}}b$
(I) $\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }}$
On rationalizing the term,
We get,
$\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} \times \dfrac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }},\,\,we\,get$
$= \dfrac{{(7 + \sqrt 5 )(7 + \sqrt 5 )}}{{(7 - \sqrt 5 )(7 + \sqrt 5 )}}$
Use the formula $\left( {a + b} \right){\text{ }}\left( {a - b} \right) = {a^2} - {b^2}$
$(7 + \sqrt 5 )(7 - \sqrt 5 ) \\ = {(7)^2} - {(\sqrt 5 )^2} \\ = 49 - 5 \\ = 44 \\$
And use formula ${(a + b)^2} = {a^2} + 2ab + {b^2}$
${\left( {7 + \sqrt 5 } \right)^2} = {(7)^2} + 2\sqrt 5 + {(\sqrt 5 )^2} \\ = 49 + 14\sqrt 5 + 5 \\$
After solving the two expressions we get,
$\dfrac{{49 + 14\sqrt 5 + 5}}{{44}} \\ = \dfrac{{54 + 14\sqrt 5 }}{{44}}....(1) \\$
Solving (II) part $\dfrac{{7\sqrt 5 }}{{7 + \sqrt 5 }}$
On rationalizing the term,
$\dfrac{{7\sqrt 5 }}{{7 + \sqrt 5 }} \times \dfrac{{7 - \sqrt 5 }}{{7 - \sqrt 5 }}$
$= \dfrac{{7\sqrt 5 (7 - \sqrt 5 )}}{{{{(7)}^2} - {{(\sqrt 5 )}^2}}} \\ = \dfrac{{49\sqrt 5 - 35}}{{49 - 5}} \\ = \dfrac{{49\sqrt 5 - 35}}{{44}} \\$
On adding two results in $\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7\sqrt 5 }}{{7 + \sqrt 5 }}$ we get,
$= \dfrac{{54 + 14\sqrt 5 - \left( {49\sqrt 5 - 35} \right)}}{{44}}$
$= \dfrac{{89 - 35\sqrt 5 }}{{44}}$
Firstly solve L.H.S by taking L.C.M
$\dfrac{{(7 + \sqrt 5 )(7 + \sqrt 5 ) - \sqrt 5 (7 - \sqrt 5 )}}{{(7 - \sqrt 5 )(7 + \sqrt 5 )}}$
$Use{\text{ }}\left( {a + b} \right){\text{ }}\left( {a - b} \right) = {a^2} - {b^2},{(a + b)^2} = {a^2} + 2ab + {b^2} \\ (7 + \sqrt 5 )(7 - \sqrt 5 ) = {(7)^2} - {(\sqrt 5 )^2} \\ = 49 - 5 \\ = 44 \\$
$\dfrac{{{{(7)}^2} + 2 \times 7 \times \sqrt 5 + {{(5)}^2} - (49\sqrt 5 - 35)}}{{49 - 5}}$
$\dfrac{{49 + 14\sqrt 5 + 5 - 49\sqrt 5 - 35}}{{44}} \\ \dfrac{{89 - 35\sqrt 5 }}{{44}} \\$
Now substitute the value of LHS which we have solved in equation (i)
$\dfrac{{89 - 35\sqrt 5 }}{{44}} = a + \dfrac{7}{{11}}\sqrt 5 b \\ or \\ \dfrac{{89}}{{44}} - \dfrac{{35}}{{44}}\sqrt 5 = a + \dfrac{7}{{11}}\sqrt 5 b \\$
On equating the coefficients of like terms, we get,
$a = \dfrac{{89}}{{44}},\dfrac{{7b}}{{11}} = \dfrac{{ - 35}}{{44}} \\ a = \dfrac{{89}}{{44}},b = \dfrac{{ - 35 \times 11}}{{7 \times 44}} \\ a = \dfrac{{89}}{{44}},b = \dfrac{{ - 5}}{4} \\$

Note: By taking LCM and using formula, we can easily find out the values. Special attention should be given to the signs while expansion of formula.