Question

if X=(2+root5)^1/2+(2-root)^1/2 and y=(2+root5)^1/2-(2-root5)^1/2 then evaluate x^2 +y^2 solve on a paper please ^means raise to

Answer 1

Question :

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

Then evaluate $x{}^{2}+y{}^{2}$

Formula used :

$1)(x + y) {}^{2} = x {}^{2} + y {}^{2} - 2xy$

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

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

Solution :

we to find the value of $x{}^{2}+y{}^{2}$

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

Now squaring on both sides

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

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

$x{}^{2}= 4 + 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} } ...(1)$

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

Squaring on both sides

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

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

$y{}^{2} = 4 - 2 \times \sqrt{2 + \sqrt{5} } \times \sqrt{2 - \sqrt{5} } ...(2)$

___________________________

Now add equation (1)&(2)

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

$\implies \: x {}^{2} + y {}^{2} = 8$

which is the required solution!

Answer 2

Step-by-step explanation:

from Dear Gyaan sir on yt