Question

How do you solve `\frac { 3} { x ^ { 2} - 6x + 8} \leq \frac { 6} { x ^ { 2} - 16}`?

Answer 1

The solution is `x in ]-oo,-4[uu]-2,4[uu[8,+oo[`

Explanation:

We cannot do crossing over

`3/((x-4)(x-2))-6/((x+4)(x-4))<=0`

`(3(x+4)-6(x-2))/((x+4)(x-4)(x+2))<=0`

`(3x+12-6x+12)/((x+4)(x-4)(x+2))<=0`

`(24-3x)/((x+4)(x-4)(x+2))<=0`

`(3(8-x))/((x+4)(x-4)(x+2))<=0`

Let `f(x)=(3(8-x))/((x+4)(x-4)(x+2))`

We can build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-4` #color(white)(aa aaa)# `-2` `color(white)(aaaaaaaa)` `4` `color(white)(aaaaaa)` `8` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+4` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `||` `color(white)(aaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `+`

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

`color(white)(aaaa)` `x-4` `color(white)(aaaa)` `-` `color(white)(aaa)` `||` `color(white)(aa)` `-` `color(white)(aaa)` `||` `color(white)(aaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `8-x` `color(white)(aaaa)` `+` `color(white)(aaa)` `||` `color(white)(aa)` `+` `color(white)(aaa)` `||` `color(white)(aaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aaaa)` `-`

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

Therefore,

`f(x)<=0` when `x in ]-oo,-4[uu]-2,4[uu[8,+oo[`