Solve the inequality (9x^2 - 64) / (x^2-25) > 0

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Solve the inequality (9x^2 - 64) / (x^2-25) > 0

Mathematics
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This ones a good one
mmm unknown the denominator , must multiply both sides by a square!
so we get (9x^2 - 64 ) ( x^2 -25)^2 > 0 now lets break it down using difference of sqaures

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Other answers:

(3x-8)(3x+8) [ (x-5)(x+5)]^2 >0 now distribute the square on the last terms ( so to speak )
(3x-8)(3x+8) (x-5)^2(x+5)^2 > 0 soln is x<-5 , -5 5 the union of those 4 regions, but I dont really know /like set notation
once 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
What would that solution look like in interval notation?
The 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
somebody dig us out of this hole!! lol
I figured out where I noobed up
when I multiplied top and bottom by the sqaure , I didnt cancel a factor on the LHS
ie it should be (9x^2 - 64 ) ( x^2 -25) > 0 (3x-8)(3x+8)(x-5)(x+5) >0 so x<-5 , -8/35 now its correct
Ah so, now x=0 is a possibility.
what happened to the denominator?
Going to take a shot. The fraction has to be greater than 0. Hence x>0 and not = to 5 Refer to the plot attachment
1 Attachment
Nice 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
*numerator and denominator can never be of opposite signs
Now my question is to elecengineer, did you multiply the LHS and the RHS by (x^2-25) and that is why the denominator disappeared?
no, 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
OK. It's got me to thinking lol.
\[x>-\frac{8}{3} \text{and } x<\frac{8}{3}\text{or } x>5\] The solution to the fraction set to zero is +-8/3
\[\left(x>-\frac{8}{3} \text{and } x<\frac{8}{3}\right)\text{or } (x<-5) \text{ or }( x>5) \]

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