Math, asked by kamalhajare543, 2 days ago

Solve This:-

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


Find:
 \sf \: Value \:  of  \:  \longrightarrow \:  \sf{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}}

Answers

Answered by user0888
45

\Large\text{\underline{\underline{What's important}}}

Even if we change x,y into y,x, the expression is equivalent to the first expression. So, this expression is symmetric.

In this case, we can evaluate with the sum and product.

\Large\text{\underline{\underline{Explanation}}}

As we know that -

\text{$\cdots\longrightarrow x^{2}+xy+y^{2}=(x+y)^{2}-xy$}

and, -

\text{$\cdots\longrightarrow x^{2}-xy+y^{2}=(x+y)^{2}-3xy$}

we have to evaluate -

\text{$\cdots\longrightarrow\boxed{\dfrac{(x+y)^{2}-xy}{(x+y)^{2}-3xy}.}$}

Let's rationalize -

\text{$\cdots\longrightarrow x=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{8-2\sqrt{15}}{2}=\boxed{4-\sqrt{15}.}$}

And then -

\text{$\cdots\longrightarrow y=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{8+2\sqrt{15}}{2}=\boxed{4+\sqrt{15}.}$}

Now we know that -

\text{$\cdots\longrightarrow\begin{cases} & x+y=8 \\  & xy=1. \end{cases}$}

The given fraction is -

\text{$\cdots\longrightarrow\boxed{\begin{aligned}&\dfrac{(x+y)^{2}-xy}{(x+y)^{2}-3xy}\\\\&=\dfrac{8^{2}-1}{8^{2}-3}\\\\&=\dfrac{63}{61}.\end{aligned}}$}

Hence, the answer is -

\text{$\cdots\longrightarrow\boxed{\dfrac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}=\dfrac{63}{61}.}$}

Answered by Dalfon
115

Answer:

63/61

Step-by-step explanation:

x = (√5 - √3)/(√5 + √3)

Rationalising the denominator,

x = (√5 - √3)/(√5 + √3) × (√5 - √3)/(√5 - √3)

x = [√5(√5 - √3) - √3(√5 - √3)]/(5 - 3)

x = (5 - √15 - √15 + 3)/2

x = (8 - 2√15)/2

x = 4 - √15

y = (√5 + √3)/(√5 - √3)

Rationalising the denominator,

y = (√5 + √3)/(√5 - √3) × (√5 + √3)/(√5 +√3)

y = [√5(√5 + √3) + √3(√5 + √3)]/(5 - 3)

y = (5 + √15 + √15 + 3)/2

y = (8 + 2√15)/2

y = 4 + √15

Now,

→ (x² + xy + y²)/(x² - xy + y²)

Used identity: (a + b)² = a² + b² + 2ab

So, we can write (x² + xy + y²)/(x² - xy + y²) as (x² + y² + 2xy - xy)/(x² + y² + 2xy - 3xy) or [(x + y)² - xy]/[(x + y)² - 3xy].

→ [(x + y)² - xy]/[(x + y)² - 3xy]

Substitute the values,

→ [(4 - √15 + 4 + √15)² - (4 - √15)(4 + √15)]/[(4 - √15 + 4 + √15)² - 3(4 - √15)(4 + √15)]

→ [(8)² - (16 + 4√15 - 4√15 - 15)]/[(8)² - 3(16 - 15)]

→ (64 - 1)/(64 - 3)

→ 63/61

Hence, the value of (x² + xy + y²)/(x² - xy + y²) is 63/61.

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