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

1 Answer
Aug 3, 2017

Answer:

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]}