Asked by Gulshad Khan
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
Answers
Answered by
Reiny
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
<b>x^2 + y^2</b>
= (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
= <b>28</b>
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
<b>x^2 + y^2</b>
= (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
= <b>28</b>
Answered by
Reiny
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)
= (x+y)^ - 2xy
about 1/2 way down the solution
should have been:
= (x+y)^2 - 2xy
(no effect on the answer)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.