Question

a=equal to root 5 +1 up on root 5 minus 1 and b=root5-1 upon root 5+1 then find a2+ab+b2/a2-ab+b2

Answer 1

check the attachment please

Answer 2

$We \:have \: a = \frac{ (\sqrt{5} + 1 ) }{(\sqrt{5} - 1 )} \\and\: b = \frac{ (\sqrt{5} -1 ) }{(\sqrt{5} + 1 )}$

$i ) a + b \\= </p><p>\frac{ (\sqrt{5} + 1 ) }{(\sqrt{5} - 1 )} + \frac{ (\sqrt{5} - 1 ) }{(\sqrt{5} +1 )}$

$= \frac{ (\sqrt{5} + 1 )^{2} + ( \sqrt{5} - 1)^{2}}{(\sqrt{5} + 1 ) ( \sqrt{5} - 1) }$

$= \frac{2[( \sqrt{5})^{2}+1^{2})]}{(\sqrt{5})^{2} - 1^{2} }$

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By Algebraic Identities:

$\pink { i ) (a+b)^{2} + (a-b)^{2} = 2(a^{2} +b^{2} )}$

$\blue { ii) (a+b)(a-b)^{2} = a^{2} - b^{2}}$

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$= \frac{ 2 ( 5 + 1 )}{5-1}\\= \frac{ 2 \times 6}{4} \\= 3 \: ---(1)$

$ii) a b \\= </p><p>\frac{ (\sqrt{5} + 1 ) }{(\sqrt{5} - 1 )} \times \frac{ (\sqrt{5} - 1 ) }{(\sqrt{5} +1 )} \\= 1 \: --(2)$

$Now, \red{ Value \: of \: \frac{ a^{2} + ab + b^{2}}{a^{2}-ab +b^{2}} }$

$= \frac{( a^{2} + 2ab + b^{2}) - ab }{(a^{2}+2ab +b^{2}) - 3ab }$

$= \frac{ (a+b)^{2} - ab }{ (a+b)^{2} - 3ab }$

$= \frac{ 3^{2} - 1 }{3^{2} - 3 } \: [ From \: (1) \:and \:(2) ] \\= \frac{ 9 - 1 }{ 9 - 3 } \\= \frac{8}{6} \\= \frac{ 4}{3}$

Therefore.,

$\red{ Value \: of \: \frac{ a^{2} + ab + b^{2}}{a^{2}-ab +b^{2}} }\green {=\frac{ 4}{3}}$

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