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24 Jun 2016, 08:21
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A
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:
31% (01:55) correct
69%
(01:52)
wrong
based on 1147
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Not Attempted Yet
What is the value of y?
(1) \(3y-1=\sqrt{(8y^2-4y+9)}\)
(2) \(y^2–2y–8 = 0\)
(1) \(3y-1=\sqrt{(8y^2-4y+9)}\)
(2) \(y^2–2y–8 = 0\)
GGMU
Current Student
Joined: 04 Feb 2014
Last visit: 11 Aug 2018
Posts: 183
Given Kudos: 164
Location: India
Concentration: General Management, Entrepreneurship
Schools: (D) XLRI (II) Alberta"20 HHL Leipzig"20 (A) (A$) GMBA '19 (WD) Smurfit (D) MBS '20 (A) Sauder '20 (S) Ryerson"19 (S) ISB '19 (D)
GPA: 3
WE:Project Management (Manufacturing)
24 Jun 2016, 09:20
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Yes the ans is A... This is a brilliant question from Magoosh... Below is the OE..
Using a tried and true DS strategy, start with the easier statement, Statement #2.
Statement #2:
(y – 4)(y + 2) = 0
y = +4 or y = –2
Since there are two values of y, this statement, alone and by itself, is not sufficient.
Statement #1:
This is an equation with a radical. The radical is already isolated, so square both sides.
(3y – 1)2 = 8y2 – 4y + 9
9y2 – 6y + 1 = 8y2 – 4y + 9
y2 – 2y – 8 = 0
Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.
Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS
Test y = –2 on equation 3y-1=√(8y^2-4y+9)
The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.
(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)
Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.
Using a tried and true DS strategy, start with the easier statement, Statement #2.
Statement #2:
(y – 4)(y + 2) = 0
y = +4 or y = –2
Since there are two values of y, this statement, alone and by itself, is not sufficient.
Statement #1:
This is an equation with a radical. The radical is already isolated, so square both sides.
(3y – 1)2 = 8y2 – 4y + 9
9y2 – 6y + 1 = 8y2 – 4y + 9
y2 – 2y – 8 = 0
Lo and behold! We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = –2. The naïve conclusion would be—this statement says exactly the same thing as the other. That's incorrect, though, because we don't know whether both of these values are valid solutions, or whether one or more is an extraneous root. We need to test this in the original equation.
Test y = +4 on equation 3y-1=√(8y^2-4y+9)
LHS=RHS
Test y = –2 on equation 3y-1=√(8y^2-4y+9)
The LHS and RHS are not equal, so this does not check! This value, y = –2, is an extraneous root.
(NB: it's often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)
Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question. This statement, alone and by itself, is sufficient.
General Discussion
24 Jun 2016, 21:02
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anurag16
Hi ,
Given : y=-2
placing the values you get : -7 = sqrt(49) , this is possible
y = 4 you get 11 = sqrt(121)
So how can statement A be sufficient ??
