Question

How do you solve and write the following in interval notation: `5(x^2-1)>24x`?

Answer 1

`x>5 or x<-1/5`

Explanation:

`5(x^2-1)>24x`
`= 5x^2-24x-5>0`

`x_(1,2)=(24+-sqrt(576+100))/10=(24+-26)/10`
`=>`
`x_1=(24+26)/10=5`
`x_2=(24-26)/10=-1/5`

`5x^2-24x-5>0`
`=> (x-5)(x+1/5)>0`

we will check for `x=0`:
`(-5)(+1/5)=(-)<0`

`=>`
`x>5 or x<-1/5`