[tex]\large\underline{\sf{Solution-}} \: [/tex]
Given that,
[tex]\sf \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a\sqrt{5} - b \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \frac{3 - 2 \sqrt{5} }{3 - 2\sqrt{5}} = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{(3 - \sqrt{5})(3 - 2\sqrt{5}) }{(3 + 2 \sqrt{5} )(3 - 2\sqrt{5})} = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{3(3 - 2\sqrt{5}) - \sqrt{5} (3 - 2\sqrt{5})}{ {3}^{2} - (2 \sqrt{5} )^{2} } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\boxed{ \bf{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{9 - 6\sqrt{5} - 3\sqrt{5} + 10}{ 9 - 20 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{19 - 9\sqrt{5}}{- 11 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{9\sqrt{5} - 19}{11 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{9\sqrt{5} }{11 } - \frac{19}{11} = a\sqrt{5} - b \\ \\ [/tex]
On comparing, we get
[tex]\sf \: \implies \: \boxed{ \sf{ \:a = \frac{9}{11} \: }} \: \: \: and \: \: \: \boxed{ \sf{ \:b = \frac{19}{11} \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Verified answer
[tex]\large\underline{\sf{Solution-}} \: [/tex]
Given that,
[tex]\sf \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a\sqrt{5} - b \\ \\ [/tex]
On rationalizing the denominator, we get
[tex]\sf \: \dfrac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \frac{3 - 2 \sqrt{5} }{3 - 2\sqrt{5}} = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{(3 - \sqrt{5})(3 - 2\sqrt{5}) }{(3 + 2 \sqrt{5} )(3 - 2\sqrt{5})} = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{3(3 - 2\sqrt{5}) - \sqrt{5} (3 - 2\sqrt{5})}{ {3}^{2} - (2 \sqrt{5} )^{2} } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\boxed{ \bf{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \: }} \\ \\ [/tex]
[tex]\sf \: \dfrac{9 - 6\sqrt{5} - 3\sqrt{5} + 10}{ 9 - 20 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{19 - 9\sqrt{5}}{- 11 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{9\sqrt{5} - 19}{11 } = a\sqrt{5} - b \\ \\ [/tex]
[tex]\sf \: \dfrac{9\sqrt{5} }{11 } - \frac{19}{11} = a\sqrt{5} - b \\ \\ [/tex]
On comparing, we get
[tex]\sf \: \implies \: \boxed{ \sf{ \:a = \frac{9}{11} \: }} \: \: \: and \: \: \: \boxed{ \sf{ \:b = \frac{19}{11} \: }} \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]