Question

Solve the inequality 30/x-1 < x+2?

Answer 1

`x\in(\frac{-1-\sqrt{129}}{2}, 1)\cup(\frac{-1+\sqrt{129}}{2}, \infty)`

Explanation:

`\frac{30}{x-1}< x+2`

`\frac{30}{x-1}-(x+2)<0`

`\frac{30-(x+2)(x-1)}{x-1}<0`

`\frac{30-x^2-x+2}{x-1}<0`

`\frac{-x^2-x+32}{x-1}<0`

`\frac{x^2+x-32}{x-1}>0`

Using quadratic formula to find the roots of `x^2+x-32=0` as follows

`x=\frac{-1\pm\sqrt{1^2-4(1)(-32)}}{2(1)}`

`x=\frac{-1\pm\sqrt{129}}{2}`

`\therefore \frac{(x+\frac{1+\sqrt{129}}{2})(x+\frac{1-\sqrt{129}}{2})}{x-1}>0`

Solving above inequality, we get

`x\in(\frac{-1-\sqrt{129}}{2}, 1)\cup(\frac{-1+\sqrt{129}}{2}, \infty)`

Answer 2

`color(blue)((-1/2-1/2sqrt(129),1)uuu(-1/2+1/2sqrt(129),oo)`

Explanation:

`30/(x-1)< x+2`

subtract `(x+2)` from both sides:

`30/(x-1)-x-2<0`

Simplify `LHS`

`(-x^2-x+32)/(x-1)<0`

Find roots of numerator:

`-x^2-x+32=0`

By quadratic formula:

`x=(-(-1)+-sqrt((-1)^2-4(-1)(32)))/(2(-1))`

`x=(1+-sqrt(129))/-2`

`x=-1/2+1/2sqrt(129)`

`x=-1/2-1/2sqrt(129)`

For `x> -1/2+1/2sqrt(129)`

`-x^2-x+32 < 0`

For `x< -1/2+1/2sqrt(129)`

`-x^2-x+32 > 0`

For `x> -1/2-1/2sqrt(129)`

`-x^2-x+32 > 0`

For `x< -1/2-1/2sqrt(129)`

`-x^2-x+32 < 0`

Root of `x-1`

`x-1=0=>x=1`

For: `x > 1`

`x-1>0`

For `x < 1`

`x-1 < 0`

Check for:

`+/-`, `-/+`

This gives us:

`-1/2-1/2sqrt(129)< x <1`

`-1/2+1/2sqrt(129)< x < oo`

In interval notation this is:

`(-1/2-1/2sqrt(129),1)uuu(-1/2+1/2sqrt(129),oo)`