Question

How do you solve and write the following in interval notation: `x^2 + x >3`?

Answer 1

The solution is `x in (-oo,(-1-sqrt13)/2 ) uu((-1+sqrt13)/2,+oo)`

Explanation:

The inequality is `x^2+x>3`

`x^2+x-3>0`

We need the roots of the quadratic equation

`x^2+x-3=0`

The discriminant is

`Delta=b^2-4ac=(1)^2-4*(1)*(-3)=1+12=13`

As `Delta>0`, there are 2 real roots

`x=(-b+-sqrtDelta)/(2a)=(-1+-sqrt13)/2`

Therefore,

`x_1=(-1-sqrt13)/2`

`x_2=(-1+sqrt13)/2`

Let `f(x)=x^2+x-3`

We can build the sign chart

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

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

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

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

Therefore,

`f(x)>0` when `x in (-oo,x_1) uu (x_2,+oo)`

graph{x^2+x-3 [-10, 10, -5, 5]}