Question

How do you solve and write the following in interval notation: `(x(x-2))/ (x^2-1x-30)<= 0`?

Answer 1

The solution is `x in (-5,0] uu[2,6)`

Explanation:

Let's factorise the denominator

`(x^2-x_30)=(x-6)(x+5)`

Let `f(x)=(x(x-2))/((x-6)(x+5)`

We can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaaaa)` `0` `color(white)(aaaaaaa)` `2` `color(white)(aaaaaa)` `6` `color(white)(aaaa)` `+oo`

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

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

`color(white)(aaaa)` `x-2` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `x-6` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

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

Therefore,

`f(x)>=0` when `x in (-5,0] uu[2,6)`

graph{(x(x-2))/((x-6)(x+5)) [-10, 10, -5, 4.995]}