Problem Solving

anoher one

Ok, I tried answering this... But, somehow I don't get the correct answer. This problem looks pretty straight.. But, I am still not getting the OA.

(squareroot of 2+ squareroot of 5)^2, I tried using (a+b)^2= a^2+2ab+b^2... By that, I get 2+2(2)(5)+5=17. The OA is 13. Hmm.. Interesting..

May be Stuart can help on this...

Its not 17, its 27. But the OA is 13...

smar83 wrote:Ok, I tried answering this... But, somehow I don't get the correct answer. This problem looks pretty straight.. But, I am still not getting the OA.

(squareroot of 2+ squareroot of 5)^2, I tried using (a+b)^2= a^2+2ab+b^2... By that, I get 2+2(2)(5)+5=17. The OA is 13. Hmm.. Interesting..

May be Stuart can help on this...

Your substitution is wrong.

The middle term is 2ab. In this case, a=root2 and b=root5. You've substituted a=2 and b=5.

You should have gotten:

2 + 2(root2)(root5) + 5 = 7 + 2(root10) = 7 + 2(approx3) = approximately 13

Sid

actually I found it. it's simple.
(sqrt 2 + sqrt5)^2=
2+2.^2^5+5=
7+2^10=
approxim 13 (2^10=6 approxim)

cheers
Nad

Ohhh....I took root 2 * root 5 as 10, instead of root 10. Thanks..