Question

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

Answer 1

This ones a good one

Answer 2

mmm unknown the denominator , must multiply both sides by a square!

Answer 3

so we get (9x^2 - 64 ) ( x^2 -25)^2 > 0

now lets break it down using difference of sqaures

Answer 4

(3x-8)(3x+8) [ (x-5)(x+5)]^2 >0

now distribute the square on the last terms ( so to speak )

Answer 5

(3x-8)(3x+8) (x-5)^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

Answer 6

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

Answer 7

What would that solution look like in interval notation?

Answer 8

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

Answer 9

somebody dig us out of this hole!! lol

Answer 10

I figured out where I noobed up

Answer 11

when I multiplied top and bottom by the sqaure , I didnt cancel a factor on the LHS

Answer 12

ie it should be (9x^2 - 64 ) ( x^2 -25) > 0

(3x-8)(3x+8)(x-5)(x+5) >0

so x<-5 , -8/3<x<8/3 , x>5

now its correct

Answer 13

Ah so, now x=0 is a possibility.

Answer 14

what happened to the denominator?

Answer 15

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

Answer 16

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

Answer 17

*numerator and denominator can never be of opposite signs

Answer 18

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?

Answer 19

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

Answer 20

OK. It's got me to thinking lol.

Answer 21

\[x>-\frac{8}{3} \text{and } x<\frac{8}{3}\text{or } x>5\]
The solution to the fraction set to zero is +-8/3

Answer 22

\[\left(x>-\frac{8}{3} \text{and } x<\frac{8}{3}\right)\text{or } (x<-5) \text{ or }( x>5) \]