Question

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

Answer 1

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

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 -

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

and, -

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

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 -

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

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 -

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

Answer 2

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.