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01 Apr 2015, 04:15
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (02:24) correct
35%
(02:31)
wrong
based on 497
sessions
History
Date
Time
Result
Not Attempted Yet
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =
A. 1
B. 2
C. 3
D. 4
E. 5
A. 1
B. 2
C. 3
D. 4
E. 5
06 Apr 2015, 06:01
Kudos
Bookmarks
Bunuel
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for a correct solution.
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for a correct solution.
MAGOOSH OFFICIAL SOLUTION:
Attachment:
rootaplusrootb_text.png [ 12.94 KiB | Viewed 23371 times ]
General Discussion
01 Apr 2015, 04:29
Kudos
Bookmarks
Bunuel
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for a correct solution.
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for a correct solution.
\(16^2\)=256
17+2\(\sqrt{66}\)=a+b+2\(\sqrt{ab}\)
b=6 a=11
a-b=5
Answer: E
a+b=17
a=\(\frac{66}{b}\)
