Question

What's the answer for this question
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Answer 1

Answer:

$\boxed{\huge{a = 3 \: \: and \: \: b = 0}}$

$\bold{Given}\longrightarrow \\ \dfrac{ \sqrt{5} - 1 }{ \sqrt{5} + 1} + \dfrac{ \sqrt{5} + 1 }{ \sqrt{5} - 1} = a + b \sqrt{5}$

$\bold{To \: find}\longrightarrow \\ value \: of \: a \: and \: b$

$\bold{Concept \: used}\longrightarrow \\ 1) {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ 2) {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ 3)( {x}^{2} - {y}^{2} ) = (x + y) \: (x - y)$

$\bold{Solution}\longrightarrow \\ \dfrac{ \sqrt{5} - 1 }{ \sqrt{5} + 1 } + \dfrac{ \sqrt{5} + 1 }{ \sqrt{5} - 1} = a + b \sqrt{5} \\ = > \dfrac{ {( \sqrt{5} - 1) }^{2} + {( \sqrt{5} + 1)}^{2} }{( \sqrt{5} + 1) \: ( \sqrt{5} - 1) } = a + b \sqrt{5} \\ = > \dfrac{ {( \sqrt{5}) }^{2} + {(1)}^{2} - 2( \sqrt{5} )(1) + {( \sqrt{5} )}^{2} + {(1)}^{2} + 2( \sqrt{5} )(1) }{ {( \sqrt{5}) }^{2} - {(1)}^{2} } = a + b \sqrt{5} \\ = > \dfrac{5 + 1 - 2 \sqrt{5} + 5 + 1 + 2 \sqrt{5} }{5 - 1} \\ = > \dfrac{12}{4} = a + b \sqrt{5} \\ = > 3 = a + b \sqrt{5} \\ = > 3 + 0 \sqrt{5} = a + b \sqrt{5} \\ on \: comparing \: we \: get \\ = > a = 3 \: \: and \: \: b = 0$