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anonymous
 5 years ago
Solve the inequality (9x^2  64) / (x^225) > 0
anonymous
 5 years ago
Solve the inequality (9x^2  64) / (x^225) > 0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0mmm unknown the denominator , must multiply both sides by a square!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we get (9x^2  64 ) ( x^2 25)^2 > 0 now lets break it down using difference of sqaures

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(3x8)(3x+8) [ (x5)(x+5)]^2 >0 now distribute the square on the last terms ( so to speak )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(3x8)(3x+8) (x5)^2(x+5)^2 > 0 soln is x<5 , 5<x<8/3 , 8/3<x<5 , x> 5 the union of those 4 regions, but I dont really know /like set notation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once again I assure you the method is very easy , just that if I try to explain it over the net it will take me a very long time, and you most likely wouldnt get a thing , however in person it is much easier to explain

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What would that solution look like in interval notation?

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The answers exclude x=0, and yet we know x could =0 64/25 is positive and larger than 0 so x could be a 0

radar
 5 years ago
Best ResponseYou've already chosen the best response.0somebody dig us out of this hole!! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I figured out where I noobed up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when I multiplied top and bottom by the sqaure , I didnt cancel a factor on the LHS

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ie it should be (9x^2  64 ) ( x^2 25) > 0 (3x8)(3x+8)(x5)(x+5) >0 so x<5 , 8/3<x<8/3 , x>5 now its correct

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Ah so, now x=0 is a possibility.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0what happened to the denominator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Going to take a shot. The fraction has to be greater than 0. Hence x>0 and not = to 5 Refer to the plot attachment

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Nice drawing, I can see x can never equal + or minus 5 (division by zero) Also for the fraction to be positive or greater than zero, the numerator and can never be of opposite signs. that is before I ever attempt to solve this thing

radar
 5 years ago
Best ResponseYou've already chosen the best response.0*numerator and denominator can never be of opposite signs

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Now my question is to elecengineer, did you multiply the LHS and the RHS by (x^225) and that is why the denominator disappeared?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, you multiply both sides by (x^2 25)^2 as there is unknowns in the denominator ( and we dont know that the denominator is positive, if for example the denominator was x^2 +25 , then I could just multiply both sides by (x^2 +25 ) as that is always positive . ) then on the LHS you get (9x^2 64 ) (x^2 25)^2 / (x^2 25) and one of the factors cancels

radar
 5 years ago
Best ResponseYou've already chosen the best response.0OK. It's got me to thinking lol.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[x>\frac{8}{3} \text{and } x<\frac{8}{3}\text{or } x>5\] The solution to the fraction set to zero is +8/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left(x>\frac{8}{3} \text{and } x<\frac{8}{3}\right)\text{or } (x<5) \text{ or }( x>5) \]
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