Question

Solve the inequality/Solve for X:

`((5x+3)(3x-2))/((x^3-7)^2) \geq 0`

Answer 1

If
`x` is real, then
`(x^3-7)^2` will always be non-negative, so the sign of the numerator will determine the solution to the inequality. In order for the expression on the left hand side to be defined, we require

`(x^3-7)^2\\eq0\implies x^3\\eq7\implies x\\eq\sqrt[3]7\approx1.91`

We have

`(5x+3)(3x-2)=0\implies x=-\frac35=-0.6,x=\frac23\approx0.66`

so there are three intervals we need to check.

(1) We pick a value of
`x` from
`-\infty<x<-\frac35`, say
`x=-1`. This gives the numerator a value of
`(-5+3)(-3-2)=(-2)(-5)=10>0`.

(2) From
`-\frac35<x<\frac23`, we can pick
`x=0` and we get
`(0+3)(0-2)=-6<0`.

(3) From
`\frac23<x<\infty`, we can pick
`x=1` and get
`(5+3)(3-2)=8>0`. But remember, we can't let
`x=\sqrt[3]7`.

So the solution to the inequality is the union of the two intervals that we showed would make the numerator positive, while still avoiding making the expression undefined:

`\left(-\infty,-\frac35\right]\cup\left(\frac23,\sqrt[3]7\right)\cup\left(\sqrt[3]7,\infty\right)`