Question

How do you solve `(x+3)/(x^2-5x+6)<=0`?

Answer 1

The answer is `x in ]-oo,-3] uu ]-1,6[`

Explanation:

Let's factorise the denominator

`x^2-5x+6=(x+1)(x-6)`

Let `f(x)=(x+3)/((x+1)(x-6))`

The domain of `f(x)` is `D_f(x)=RR-{-1,6}`

Now, we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaaaa)` `-1` `color(white)(aaaaaaaaa)` `6` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-6` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0` when `x in ]-oo,-3] uu ]-1,6[`