How do you solve #x² + 2x - 15 ≤ 0#?

1 Answer
Apr 15, 2018

Answer:

The solution is #x in [-5, 3]#

Explanation:

The inequality is

#x^2+2x-15<=0#

#(x+5)(x-3)<=0#

Let #f(x)=(x+5)(x-3)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaa)##-5##color(white)(aaaaaaaa)##3##color(white)(aaaaaa)##+oo#######

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaa)##0##color(white)(aaa)##+##color(white)(aaaaaaa)##+#######

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

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

Therefore,

#f(x)<=0# when #x in [-5, 3]#

graph{x^2+2x-15 [-36.52, 36.52, -18.26, 18.28]}