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07 Apr 2023, 03:38
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Hi, I would really appreciate some help regarding the general rule to proceed in questions in which we solve for X>0, but we are unsure about the solutions for X<0. Should I solve the inequality for X<0 as well, or is it sufficient to check specific ranges of X based on the root of the expression?
An example:
IF sqrt(X+2)>X , WHAT IS THE RANGE OF POSSIBLE SOLUTIONS OF X?
So, first, since the expression within the square root has to be positive, -2<X.
Then, we must assume that X>0 to square the inequality.
The final equation is x^2-x-2<0, while -1<x<2.
But, I had to assume that X>0.
Would the next step be to:
1. Also solve the inequality for X<0 and check the solutions?
(In this case, -2<X<0 because of the square root)
OR
2. Only check if -1<X<0 is correct to see if It should be included in the answer, and then check for -2<X<-1 because of the square root?
Thank you!!
An example:
IF sqrt(X+2)>X , WHAT IS THE RANGE OF POSSIBLE SOLUTIONS OF X?
So, first, since the expression within the square root has to be positive, -2<X.
Then, we must assume that X>0 to square the inequality.
The final equation is x^2-x-2<0, while -1<x<2.
But, I had to assume that X>0.
Would the next step be to:
1. Also solve the inequality for X<0 and check the solutions?
(In this case, -2<X<0 because of the square root)
OR
2. Only check if -1<X<0 is correct to see if It should be included in the answer, and then check for -2<X<-1 because of the square root?
Thank you!!
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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.
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07 Apr 2023, 03:52
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HaSmamit
\(\sqrt{x+2} > x\)
There are 2 cases:
Case 1:
x is negative (x < 0)
Since LHS is always non negative (it is the principal square root so it is never negative) so this inequality will hold for all values of x.
But since under the square root, we cannot have a negative value, (x + 2) >= 0
x >= -2
Case 2:
x is 0 or positive (x >= 0)
Now we can square both sides to get
(x +1)(x - 2) < 0
So -1 < x < 2
But since x can't be negative in this case, 0 <= x < 2
Hence, overall range is -2 <= x < 2
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!
