Question

ANSWER CORRECTLY WITHOUT SPAMING​

Answer 1

Answer:

x=(√5-2)/(√5+2)

y=(√5+2)/(√5-2)

x.y=(√5-2)/(√5+2).(√5+2)/(√5-2)=1

x+y=(√5-2)/(√5+2)+(√5+2)/(√5-2)

={(√5-2)^2+(√5+2)^2}/{(√5+2).(√5-2)}

=(5+4-4√5+5+4+4√5)/(√5^2-2^2)

=18/(5-4)

=18/1

=18

so,

x^2+y^2+xy

=(x+y)^2-2xy+xy

=(x+y)^2-xy

=(18)^2-1

=324-1

=323

Answer 2

$\overline{\underline{\boxed{\sf \: QUESTION \: \darr}}}$

$\sf{If \: x = \frac{ \sqrt{5} - 2}{ \sqrt{5} + 2} \: and \: y = \frac{ \sqrt{5} + 2 }{ \sqrt{5} - 2} \: then \: find \: the \: value \: of \: {x}^{2} + xy + {y}^{2} }$

$\overline{\underline{\boxed{\sf SOLUTION \: \darr}}}$

We are given that

$\green{\sf{x = \frac{ \sqrt{5} - 2 }{ \sqrt{5} + 2}}}$

Hence, by rationalising it we get

$\sf \mapsto \blue{x = \frac{( \sqrt{5} - 2)( \sqrt{5} - 2) }{( \sqrt{5} + 2)( \sqrt{5} - 2)}}$

Applying identity :

For numerator

• (a - b)(a - b) = a² - 2ab + b²

For denominator

• (a + b)(a - b) = a² - b²

Here,

a refers √5

ab refers √5 × 2 = 2√5

b refers 2

$\sf \mapsto \blue{x = \frac{ {( \sqrt{5} - 2) }^{2} }{ {( \sqrt{5}) }^{2} - {(2)}^{2} } }$

$\sf \mapsto \blue{x = \frac{ {( \sqrt{5} )}^{2} - 2 \times \sqrt{5} \times 2 + {(2)}^{2} }{5 - 4} }$

$\sf \mapsto \blue{x = \frac{5 - 4 \sqrt{5} + 4}{1} }$

$\sf \mapsto \blue{x = 9 - 4 \sqrt{5} }$

$☢\:\red{\underline{\boxed{ \orange{\sf x = 9 - 4 \sqrt{5}}}}}$

• So after rationalising the value of x is 9 - 4√5

Now, to rationalise y

$\sf \leadsto \red{y = \frac{ \sqrt{5} + 2}{ \sqrt{5} - 2 }}$

Rationalising it we get

$\sf \leadsto \purple{y = \frac{ (\sqrt{5} + 2)( \sqrt{5} + 2)}{( \sqrt{5} - 2)( \sqrt{5} + 2)}}$

Applying identity :

For numerator

• (a + b)(a + b) = a² + 2ab + b²

For denominator

• (a + b)(a - b) = a² - b²

Here,

a refers √5

ab refers √5 × 2 = 2√5

b refers 2

$\sf \leadsto \purple{y = \frac{ {( \sqrt{5} + 2) }^{2} }{ {( \sqrt{5}) }^{2} - {(2)}^{2} } }$

$\sf \leadsto \purple{y = \frac{ {( \sqrt{5}) }^{2} + 2 \times \sqrt{5} \times 2 + {(2)}^{2} }{5 - 4} }$

$\sf \leadsto \purple{y = \frac{5 + 4 \sqrt{5} + 4 }{1} }$

$☣ \: \underline{\boxed{ \purple{ \sf y = 9 + 4 \sqrt{5} }}}$

Now to find the given condition

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

Putting the values we get :

$\sf\looparrowright\green{({9 - 4 \sqrt{5})}^{2} + (9 - 4 \sqrt{5})(9 + 4 \sqrt{5}) + {(9 + 4 \sqrt{5} )}^{2} }$

$\small\sf\looparrowright\green{ {(9)}^{2} - 72\sqrt{5} + {(4 \sqrt{5}) }^{2} + {(9)}^{2} - {( 4\sqrt{5} )}^{2} + {(9)}^{2} + 72\sqrt{5} + {(4 \sqrt{5} )}^{2} }$

$\sf\looparrowright\green{ {(9)}^{2} + {(9)}^{2} + {(9)}^{2} + {(4 \sqrt{5}) }^{2} }$

$\sf\looparrowright\green{81 + 81 + 81 + 80}$

$\sf\looparrowright\green{323}$

• Hence, the answer is 323

$\overline{ \underline{ \boxed{ \sf LEARN \: MORE \: \darr}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )}}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b)}^{2} - {(a - b)}^{2} = 4ab }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) }}}$

$\star \: \underline{ \boxed{ \mathfrak{ {(a + b - c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc - ca) }}}$