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02 Feb 2023, 02:30
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (01:47) correct
43%
(01:35)
wrong
based on 49
sessions
History
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Time
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Not Attempted Yet
Is x = 4 a solution to the equation \(x^2 + mx − 48 = 0\)?
(1) \((x+4)\) is a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
(2) \((x−4)\) is NOT a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
700-Level Question!!!
(1) \((x+4)\) is a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
(2) \((x−4)\) is NOT a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
700-Level Question!!!
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
02 Feb 2023, 07:43
Kudos
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Bunuel
If x = 4 were a root of the equation
\((4)^2 + 4m − 48 = 0\)
m = \(\frac{48 - 16}{ 4}\)
m = 8
Let's start with Statement 2 as its pretty straight-forward
Statement 2
(2) \((x−4)\) is NOT a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
If x - 4 is not a factor of the equation, x = 4 is not one of the root/solution to the equation. The statement is sufficient, and hence we can eliminate A, C, and E.
Statement 1
(1) \((x+4)\) is a factor of \(x^2 + mx − 48\), where m is a constant, and x is a variable.
As x = -4 is a solution to the equation
\((-4)^2 + (-4)m − 48\)
m = -8
Hence, m = 4 is not the root of the equation. The information is sufficient.
Option D
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