Question

How do you solve `2x+2>x^2` using a sign chart?

Answer 1

The solution is `x in (1-sqrt3, 1+sqrt3)`

Explanation:

Let's rewrite the equation

`2x+2>x^2`

`x^2-2x-2<0`

The roots of the equation

`x^2-2x-2=0`

are `x_1` and `x_2`

`x_2=(2+sqrt(2^2-(4*1*-2)))/2=(2+sqrt12)/2=(2+2sqrt3)/2=1+sqrt3`

`x_1=(2-sqrt(2^2-(4*1*-2)))/2=(2-sqrt12)/2=(2-2sqrt3)/2=1-sqrt3`

Let `f(x)=x^2-2x-2`

We build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaaa)` `x_1` `color(white)(aaaa)` `x_2` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-x_1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-x_2` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<0`, when `x in (x_1,x_2)`