Question

pls answer it .... ​

Answer 1

Answer:

Given that

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

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

We know that power 1/2 = square root

Now,

$x = \sqrt{2 + \sqrt{5} } + \sqrt{2 - \sqrt{5} }$

assuming as equation ( 1 )

$y = \sqrt{2 + \sqrt{5} } - \sqrt{2 - \sqrt{5} }$

assuming as equation ( 2 )

equation (1) squaring on both sides (S.O.B.S)

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

Using

(a + b)² = a² + b² + 2ab

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

Using

(a + b)(a - b) = a² - b²

${x}^{2} = 2 + \sqrt{5} + 2 - \sqrt{5} + 2 \sqrt{4 - 5}$

We know that

√-1 = i

Because it is a complex number

Complex number : it is a negative number present under root

${x}^{2} = 4 + 2i$

Assuming as equation (3)

Now ,

Equation ( 2 ) squaring on both sides (S.O.B.S)

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

Using (a - b)² = a² + b² - 2ab

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

${y}^{2} = 2 + \sqrt{5} + 2 - \sqrt{5} - 2i$

${y}^{2} = 4 - 2i$

Assuming as equation (4)

Equation (3) + Equation (4)

${x}^{2} + {y}^{2} = 4 + 2i + 4 - 2i$

${x}^{2} + {y}^{2} = 8$