How do you solve #x^2+5x<=0# using a sign chart?

1 Answer
Jan 13, 2017

Answer:

The answer is #=x in [-5, 0]#

Explanation:

Let's factorise the equation

#x^2+5x=x(x+5)#

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

Let's do the sign chart

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

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

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##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 [-5, 0]#