Question

How do you solve `x^{2} + 2x > 35`?

Answer 1

The solution is `x in (-oo, -7) uu(5, +oo)`

Explanation:

The inequality is

`x^2+2x>35`

`x^2+2x-35>0`

Factorise,

`(x-5)(x+7)>0`

Let `f(x)=(x-5)(x+7)`

Build a sign chart

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

`color(white)(aaaa)` `x+7` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `color(white)(a)` `+`

`color(white)(aaaa)` `x-5` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `color(white)(a)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `color(white)(a)` `+`

Therefore,

`f(x)>0` when `x in (-oo, -7) uu(5, +oo)`

graph{x^2+2x-35 [-52.25, 51.85, -39.94, 12.06]}