Question

if x =(2+√5)1/2+(2-√5)1/2 and y =(2+√5)1/2-(2-√5)1/2 then evaluate x^2+y^2

Answer 1

Answer:

Step-by-step explanation:

Using the maths property:

(a+b)(a-b) = $a^{2}-b^{2}$;

x = (2+√5)1/2+(2-√5)1/2

Taking 1/2 as common;

x = (1/2) [(2+√5)+(2-√5)] = (1/2) [2+2];

x = 2;

y = (2+√5)1/2-(2-√5)1/2;

Taking 1/2 as common;

y = (1/2) [2+√5 - 2+√5]

y = (1/2) [2√5] = √5;

x^2+y^2 = 4+5 = 9

Answer 2

X=(2+√5)1/2+(2-√5)1/2

X=(2+√5)1/2+(2-√5)1/2x^2={(2+√5)1/2+(2-√5)1/2}^2

X=(2+√5)1/2+(2-√5)1/2x^2={(2+√5)1/2+(2-√5)1/2}^2 ={√(2+√5)^2}+2*√(2+√5)*√(2-√5)+√(2-√5)^2}

=2+√5+2-√5+2*2+√5+2-√5

=4+8=12

y=(2+√5)1/2-(2-√5)1/2

y=(2+√5)1/2-(2-√5)1/2y^2={(2+√5)1/2-(2-√5)1/2}^2

y=(2+√5)1/2-(2-√5)1/2y^2={(2+√5)1/2-(2-√5)1/2}^2 ={√(2+√5)^2}-2*√(2+√5)*√(2-√5)+√(2-√5)^2}

y=(2+√5)1/2-(2-√5)1/2y^2={(2+√5)1/2-(2-√5)1/2}^2 ={√(2+√5)^2}-2*√(2+√5)*√(2-√5)+√(2-√5)^2} =2+√5+2-√5-2*2+√5+2-√5

y=(2+√5)1/2-(2-√5)1/2y^2={(2+√5)1/2-(2-√5)1/2}^2 ={√(2+√5)^2}-2*√(2+√5)*√(2-√5)+√(2-√5)^2} =2+√5+2-√5-2*2+√5+2-√5 =4-8=-4

x^2+y^2=12+(-4)=8

HOPE THIS WILL HELP YOU.