Question

Solving inequalities. How to solve `(x+5)/(3-x^2)≥ 0`?

Answer 1

See details below

Explanation:

A fraction is positive or zero if and only if numerator and denominator have the same sign

Case 1.- Both positives
`x+5>=0` then `x>=-5` and
`3-x^2>0` (imposible to be zero) then `3>x^2` that is

`-sqrt3 < x < sqrt3`

The intersection of both sets of values is `[-5,oo)nn(-sqrt3,sqrt3)=(-sqrt3,sqrt3)`

Case 2.- Both negatives

Similarly the solutions are `(-oo,-5]nn((-oo,-sqrt3)uu(sqrt3,+oo))=`

`=[-5,-sqrt3)uu(sqrt3,+oo)`

Now, the union of both cases will be the final result

`[-5,-sqrt3)uu(-sqrt3,sqrt3)uu(sqrt3,+oo)`

Answer 2

The solution is `x in (-oo,-5]uu(-sqrt3, sqrt3)`

Explanation:

The inequality is

`(x+5)/(3-x^2)>=0`

`(x+5)/((sqrt3-x)(sqrt3+x))>=0`

Let `f(x)=(x+5)/((sqrt3-x)(sqrt3+x))`

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `-sqrt3` `color(white)(aaaa)` `+sqrt3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+5` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `sqrt3+x` `color(white)(aaa)` `-` `color(white)(aaa)` ###color(white)(aaa)#`-` `color(white)(aaa)` `||` `color(white)(aa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `sqrt3-x` `color(white)(aaa)` `+` `color(white)(aaa)` ###color(white)(aaa)#`+` `color(white)(aaa)` ###color(white)(aaa)#`+` `color(white)(aa)` `||` `color(white)(aa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaa)` `0` `color(white)(aa)` `-` `color(white)(aaa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `-`

Therefore,

`f(x)>=0` when `x in (-oo,-5]uu(-sqrt3, sqrt3)`

graph{(x+5)/(3-x^2) [-12.66, 12.66, -6.33, 6.33]}