Question

pls........answer the first part of the 25th question....in the attached paper

Answer 1

Let \[(2+\sqrt{5})=a\]
Then
\[(2+\sqrt{5})^{\frac{1}{2}}=\sqrt{a}\]
Let
\[(2-\sqrt{5})=b\]
Then
\[(2-\sqrt{5})^{\frac{1}{2}}=\sqrt{b}\]
\[x ^{2}+y ^{2}=(\sqrt{a}+\sqrt{b})^{2}+(\sqrt{a}-\sqrt{b})^{2}\]
Now you should be able to find x^2 + y^2 in terms of a and b. Then substitute the values for a and b to find the answer.

Answer 2

@basith Can you work out this part?
\[(\sqrt{a}+\sqrt{b})^{2}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})=?\]

Answer 3

no,,,,,im
confused

Answer 4

Do you know how to work out the following?
\[(a+b)^{2}=(a+b)(a+b)=\]

Answer 5

\[(a+b)(a+b)=a ^{2}+2ab+b ^{2}\]
So now can you work out this part?
\[(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})=\]

Answer 6

@basith What step are you having trouble with?

Answer 7

@basith Hello!!!

Answer 8

ya

Answer 9

\[(\sqrt{a})^{2}+2\times \sqrt{a}\times \sqrt{b}+(\sqrt{b})^{2}\]

Answer 10

Well done :) And that simplifies to the following:
\[a+2\sqrt{a}\sqrt{b}+b\]
Also
\[(\sqrt{a}-\sqrt{b})^{2}=a-2\sqrt{a}\sqrt{b}+b\]
\[x ^{2}+y ^{2}=a+2\sqrt{a}\sqrt{b}+b+a-2\sqrt{a}\sqrt{b}+b=2a+2b\]
So now can you find the answer by substituting the values of a and b?

Answer 11

hw?

Answer 12

oohhh ...ya

Answer 13

\[x ^{2}+y ^{2}=2a+2b=2(2+\sqrt{5})+2(2-\sqrt{5})=4+2\sqrt{5}+4-2\sqrt{5}=?\]

Answer 14

yes.....nw i understand ...thnx a lot......and may i ask ur real name??

Answer 15

=8

Answer 16

You're welcome:)
My real name is in my profile. Just view it by clicking on my avatar and selecting 'view detailed profile'.
Good work. The correct answer is 8.