If X=(2+√5)1⁄2 + (2-√5)1⁄2 and Y=(2+√5)1⁄2 - (2-√5)1⁄2 ,the evaluate x^2+ y^2
I assume those 1/2 's are exponents so your question is
x = √(2+√5) + √(2-√5)
y = √(2+√5) - √(2-√5)
Now clearly √(2-√5) is not a real number , so this must deal with imaginary numbers.
for ease of typing let a = √2+√5 , b = √2 - √5
(where b is an imaginary number)
so x = a+b
and y = a-b
x^2 + y^2
= (x+y)^ - 2xy
= (a+b+a-b)^2 - 2(a+b)(a-b)
= (2a)^2 - 2(a^2 - b^2)
= 2a^2 + 2b^2
now a^2 = 2 + 2√10 + 5 = 7+2√10
and b^2 = 2 - 2√10 + 5 = 7 - 2√10
so
2a^2 + 2b^2
= 2(7+2√10) + 2(7-2√10)
= 14 + 4√10 + 14 - 4√10
= 28
looks like I left out the exponents in line
= (x+y)^ - 2xy
about 1/2 way down the solution
should have been:
= (x+y)^2 - 2xy
(no effect on the answer)
To evaluate X^2 + Y^2, we need to substitute the values of X and Y into the expression.
Let's start by simplifying X and Y individually:
X = (√(2+√5) + √(2-√5)) + (√(2+√5) - √(2-√5))
X = 2√(2+√5)
Y = (√(2+√5) + √(2-√5)) - (√(2+√5) - √(2-√5))
Y = 2√(2-√5)
Now, we can square both X and Y:
X^2 = (2√(2+√5))^2
X^2 = 4(2+√5)
X^2 = 8 + 4√5
Y^2 = (2√(2-√5))^2
Y^2 = 4(2-√5)
Y^2 = 8 - 4√5
Finally, we can substitute the values back into the expression:
X^2 + Y^2 = (8 + 4√5) + (8 - 4√5)
X^2 + Y^2 = 16 + 0
X^2 + Y^2 = 16
Therefore, X^2 + Y^2 evaluates to 16.