Help me please
- Thread starter Shashwat
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Dr.Peterson
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Dr.Peterson
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The equation has no real solutions (try graphing it); is the context one in which the square root of a negative or complex number is allowed?
I suspect there is an error in the problem; or perhaps it is assumed that some trick will be used that doesn't reveal that the condition is never satisfied. (I can't see one.)
Where does this problem come from?
I suspect there is an error in the problem; or perhaps it is assumed that some trick will be used that doesn't reveal that the condition is never satisfied. (I can't see one.)
Where does this problem come from?
Steven G
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Clearly Dr Peterson must be correct that there no real solution.
x>0 since we have to compute sqrt(x) in the denominator.
Both terms, 3x and 2/sqrt(x) are both positive if x>0.
The sum is 1, so 2/sqrt(x) must be less than or equal to 1. This requires sqrt(x)>2 or x>4. But then 3x would be at least 12. Hence the sum can't be 1.
x>0 since we have to compute sqrt(x) in the denominator.
Both terms, 3x and 2/sqrt(x) are both positive if x>0.
The sum is 1, so 2/sqrt(x) must be less than or equal to 1. This requires sqrt(x)>2 or x>4. But then 3x would be at least 12. Hence the sum can't be 1.