How do you solve #x^2+2x-15<=0# using a sign chart?

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Answer 1 · 2016-12-18T08:11:25

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

Let #f(x)=x^2+2x-15=(x-3)(x+5)#

Let's do the sign chart

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

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

So,

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