How do you solve the inequality #x^2+2x-3>=0# and write your answer in interval notation?

1 Answer
Aug 14, 2017

Answer:

The solution is #x in (-oo, -3] uu [1, +oo)#

Explanation:

Let's factorise the inequality

#x^2+2x-3>=0#

#(x+3)(x-1)>=0#

Let #f(x)=(x+3)(x-1)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##-3##color(white)(aaaaaaaaa)##1##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaa)##0##color(white)(aaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaa)####color(white)(aaaaa)##-##color(white)(aaa)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaa)##0##color(white)(aaaa)##-##color(white)(aaa)##0##color(white)(aa)##+#

Therefore,

#f(x)>=0# when #x in (-oo, -3] uu [1, +oo)#

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