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Solve the inquality: (3-x) / (15-x)^1/2 The solution has to be of form (a,b)
(3-x) / (15-x)^1/2 <1 (3-x) <(15-x)^1/2 (3-x)2 <|15-x| case 1: x<15 (3-x)2 <15-x therefore we get by solving this equation x = (6,-1) case 1: x>15 (3-x)2 <x-15 therefore we get, x=(7+145^1/2)/2 , (7-145^1/2)/2 but both are less than 15, thus no solution from this case thus final answer is (6,-1)
(3-x) / (15-x)^1/2 <1
(3-x) <(15-x)^1/2
(3-x)2 <|15-x|
case 1: x<15
(3-x)2 <15-x
therefore we get by solving this equation
x = (6,-1)
case 1: x>15
(3-x)2 <x-15
therefore we get,
x=(7+145^1/2)/2 , (7-145^1/2)/2
but both are less than 15,
thus no solution from this case
thus final answer is (6,-1)
I would like to point out that necessarily 15-x>0. So the case x>15 is redundant
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