Answer
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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.
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.
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