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22 Feb 2025, 01:02
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x: 6
y: -3
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
56% (02:36) correct
44%
(02:37)
wrong
based on 312
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Not Attempted Yet
The number \(\sqrt{63-36√3}\) can be written in the form of x+y√3 for some integers x and y. Choose the values of x and y in the table which are consistent with the given information
| x | y | |
| -6 | ||
| -3 | ||
| 0 | ||
| 2 | ||
| 3 | ||
| 6 |
ShowHide Answer
Official Answer
x: 6
y: -3
23 Feb 2025, 10:39
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Let's simplify \(\sqrt{63-36√3}\) to get to the form x+y√3
= \(\sqrt{9(7-4√3)}\)
= \(3\sqrt{(3+4-4√3)}\)
= \(3\sqrt{(2-√3)^2}\)
= \(3(2-√3)\) (Given 2>√3 and square root gives a positive number, so 2-√3>0)
= \(6-3√3\)
x = 6, y = -3
= \(\sqrt{9(7-4√3)}\)
= \(3\sqrt{(3+4-4√3)}\)
= \(3\sqrt{(2-√3)^2}\)
= \(3(2-√3)\) (Given 2>√3 and square root gives a positive number, so 2-√3>0)
= \(6-3√3\)
x = 6, y = -3
GMATinsight
General Discussion
Given that \(\sqrt{63-36√3}\) = x+y√3 .
Square both sides -> \(\sqrt{63-36√3}^2 = (x+y√3)^2 \)
=> 63-36√3 = \( x^2 + 3y^2 + 2√3xy \)
=> \( x^2 + 3y^2 = 63 \) and 2√3xy= -36√3
=> xy = -18 (possible options for x and y are 9,-2, ; 6,-3 ; 18,-1)
Please note that x needs to be positive and greater than y for the term x+y√3 to be positive. This term can't be negative since it is the square root of 63-36√3.
The only values that satisfy both equations xy = -18 and \( x^2 + 3y^2 = 63 \) are x = 6 and y = -3.
Hope this helps!
Square both sides -> \(\sqrt{63-36√3}^2 = (x+y√3)^2 \)
=> 63-36√3 = \( x^2 + 3y^2 + 2√3xy \)
=> \( x^2 + 3y^2 = 63 \) and 2√3xy= -36√3
=> xy = -18 (possible options for x and y are 9,-2, ; 6,-3 ; 18,-1)
Please note that x needs to be positive and greater than y for the term x+y√3 to be positive. This term can't be negative since it is the square root of 63-36√3.
The only values that satisfy both equations xy = -18 and \( x^2 + 3y^2 = 63 \) are x = 6 and y = -3.
Hope this helps!
