Question

How do you solve `(x^2+2x-15)/(x^2+7x)>0`?

Answer 1

The answer is `x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[`

Explanation:

Let `f(x)=(x^2+2x-15)/(x^2+7x)`

`=((x-3)(x+5))/(x(x+7))`

The domain of `f(x)` is `D_f(x)=RR-{0,-7}`

To solve the equation `f(x)>0`, we need a sign chart.

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-7` `color(white)(aaaa)` `-5` `color(white)(aaaa)` `0` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+7` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x+5` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaa)` `+`

So,

`f(x)>0`

when `x in ] -oo,-7 [uu ] -5, 0[ uu ]3, oo[`

graph{(x^2+2x-15)/(x^2+7x) [-16.02, 16.01, -8.01, 8.01]}