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22 Apr 2019, 14:45
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1) -2x >0
2) x^(3)<0
Didn't want to post this question in the DS forum because I have a question about this...
If this is dealing with inequalities can I isolate for x in the above DS options to determine if x is indeed less than zero? I don't really have to test cases, do I?
Thanks!
2) x^(3)<0
Didn't want to post this question in the DS forum because I have a question about this...
If this is dealing with inequalities can I isolate for x in the above DS options to determine if x is indeed less than zero? I don't really have to test cases, do I?
Thanks!
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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22 Apr 2019, 18:53
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No, you don't have to test numbers in such a simple question. The question is testing the basic rules of inequalities.
1) -2x > 0
Divide both sides by positive 2
-x > 0
Multiply both sides by -1
x < 0
Note: Whenever you multiply an inequality with a negative number, the sign reverses.
2) \(x^{3}\)< 0
= \(x^{2}\) X \(x^{1}\) < 0
Now \(x^{2}\) is always positive, so we can exclude the expression
\(x^{1}\) < 0
x < 0
Therefore, each of the statements says that x < 0, i.e. x is a negative number.
1) -2x > 0
Divide both sides by positive 2
-x > 0
Multiply both sides by -1
x < 0
Note: Whenever you multiply an inequality with a negative number, the sign reverses.
2) \(x^{3}\)< 0
= \(x^{2}\) X \(x^{1}\) < 0
Now \(x^{2}\) is always positive, so we can exclude the expression
\(x^{1}\) < 0
x < 0
Therefore, each of the statements says that x < 0, i.e. x is a negative number.
22 Apr 2019, 23:46
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Yes, you do not need to test specific cases as this is a simple inequalities problem.
Considering the two inequalities:
1. -2x > 0.
Taking LHS to RHS, we get:
0 > 2x. Dividing by 2 on both sides:
0 > x --> x < 0
2. x^(3) < 0.
Consider x^(3) like x . x^(2). x^2 is always positive (as it is a square). Hence, the sign will depend on x.
Hence, if x^(3) is less than zero then x must be less than zero.
Both statements independently prove that x < 0.
Considering the two inequalities:
1. -2x > 0.
Taking LHS to RHS, we get:
0 > 2x. Dividing by 2 on both sides:
0 > x --> x < 0
2. x^(3) < 0.
Consider x^(3) like x . x^(2). x^2 is always positive (as it is a square). Hence, the sign will depend on x.
Hence, if x^(3) is less than zero then x must be less than zero.
Both statements independently prove that x < 0.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
