Question

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Answer 1

Given:

$\longrightarrow\mathrm{x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}}$

$\longrightarrow\mathrm{y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

To Find:

Value of  $\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

Solution:

We know that,

$\longrightarrow\bf{(a+b)^{2}=a^{2}+2ab+b^{2}}$

$\longrightarrow\bf{(a-b)^{2}=a^{2}-2ab+b^{2}}$

$\longrightarrow\bf{(a+b)(a-b)=a^{2}-b^{2}}$

Now,

$\mathrm{x+y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x+y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}+(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x+y=\dfrac{5+3-2\sqrt{5}\sqrt{3}+5+3+2\sqrt{5}\sqrt{3}}{5-3}}$

$\mathrm{x+y=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}}$

$\mathrm{x+y=\dfrac{16}{2}}$

$\mathrm{x+y=8}$

Now again,

$\mathrm{x-y=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{x-y=\dfrac{(\sqrt{5}-\sqrt{3})^{2}-(\sqrt{5}+\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-(5+3+2\sqrt{5}\sqrt{3})}{5-3}}$

$\mathrm{x-y=\dfrac{5+3-2\sqrt{5}\sqrt{3}-5-3-2\sqrt{5}\sqrt{3}}{2}}$

$\mathrm{x-y=\dfrac{8-2\sqrt{15}-8-2\sqrt{15}}{2}}$

$\mathrm{x-y=\dfrac{-4\sqrt{15}}{2}}$

$\mathrm{x-y=-2\sqrt{15}}$

Also,

$\mathrm{xy=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}}$

$\mathrm{xy=1}$

So,

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}$

On adding and subtracting xy in numerator and denominator, we get

$\rightarrow\sf{\dfrac{x^{2}+xy+y^{2}+xy-xy}{x^{2}-xy+y^{2}+xy-xy}}$

$\rightarrow\sf{\dfrac{x^{2}+2xy+y^{2}-xy}{x^{2}-2xy+y^{2}+xy}}$

$\rightarrow\sf{\dfrac{(x+y)^{2}-xy}{(x-y)^{2}+xy}}$

On feeding values in above equation, we get

$\rightarrow\sf{\dfrac{(8)^{2}-1}{(-2\sqrt{15})^{2}+1}}$

$\rightarrow\sf{\dfrac{64-1}{60+1}}$

$\rightarrow\sf{\dfrac{63}{61}}$

Hence, the required value is  $\bf{\dfrac{63}{61}}$.

Answer 2

Given :

$\sf x = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$ and $\sf y = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

To find :

The value of

$\sf\dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} }$

Formula's used:

$\sf1)(x + y)(x - y) = x {}^{2} - y {}^{2}$

$2) \sf(x + y) {}^{2} = x {}^{2} + y {}^{2} + 2xy$

$3) \sf(x -y) {}^{2} = x {}^{2} + y {}^{2} - 2xy$

Solution :

$\sf x = \dfrac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

Now rationalise the denominator

$\sf x = \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

$\implies \sf x = \frac{( \sqrt{5} - \sqrt{3} ) {}^{2} }{ (\sqrt{5} + \sqrt{3} )( \sqrt{5} - \sqrt{3} ) }$

We know that (a+b)(a-b)=a²-b²

$\implies \sf x = \frac{( \sqrt{5} - \sqrt{3} ) {}^{2} }{( \sqrt{5}) {}^{2} - ( \sqrt{3} ) {}^{2} }$

we know that (a-b)²= a ² +b²-2ab

$\implies \sf x = \frac{5 + 3 - 2 \sqrt{3} \times \sqrt{5} }{2}$

$\implies \sf x = \frac{8 - 2 \sqrt{15} }{2}$

$\implies \sf x = 4 - \sqrt{15} ...(1)$

$\sf y = \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

Rationalise the denominator

$\implies \sf y = \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$\sf \implies y = \frac{( \sqrt{5} + \sqrt{3}) {}^{2} }{2}$

$\implies \sf y = \frac{5 + 3 + 2 \sqrt{15} }{2}$

$\implies \sf y = \frac{8 + 2 \sqrt{15} }{2}$

$\implies \sf y = 4 + \sqrt{15} ...(2)$

We have to find the value of

$\sf \dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} }$

First solve Numerator

$\sf x {}^{2} + xy + y {}^{2}$

Put the values of x and y from equation (1) and (2)

$= \sf (4 - \sqrt{15}) {}^{2} + (4 - \sqrt{15} )(4 + \sqrt{15} ) + (4 + \sqrt{15} ) {}^{2}$

$\sf = (16 + 15 - 2 \times 4 \times \sqrt{15} ) + (4 {}^{2} - (\sqrt{15}) {}^{2} ) + (16 + 15 + 2 \times 4 \times \sqrt{15} )$

$= 31 - \cancel{8 \sqrt{15} } - 1 + 31 + \cancel{8 \sqrt{15} }$

$\sf = 62 + 1 = 63$

Now solve denominator

$\sf x {}^{2} - xy + y {}^{2}$

Put the values of x and y from equation (1) and (2)

$\sf = (4 - \sqrt{15}) {}^{2} - (4 {}^{2} - ( \sqrt{15}) {}^{2} ) + (4 + \sqrt{3}) {}^{2}$

$\sf = 16 + 15 - 8 \sqrt{15} - (16 - 15) + 16 + 15 + 8 \sqrt{15}$

$\sf = 31 - \cancel{8 \sqrt{15} } - 1 + 31 + \cancel{8 \sqrt{15}}$

$\sf = 62 - 1 = 61$

Therefore,

$\sf \dfrac{x {}^{2} + xy + y {}^{2} }{x {}^{2} - xy + y {}^{2} } = \dfrac{63}{61}$

It is the required solution!