How do solve #(x^2+5x)/(x-3)>=0# and write the answer as a inequality and interval notation?

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Answer 1 · 2016-11-20T15:20:46

The answer is #x in [-5,0 ] uu ]3, +oo[ #

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

The domain of #f(x)# is #D_(f(x))=RR-{3}#

Let's do a sign chart to solve this inequality

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##0##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##+##color(white)(aaa)##+#

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

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

Therefore #f(x)>=0# when #x in [-5,0 ] uu ]3, +oo[#

graph{(x^2+5x)/(x-3) [-72.6, 75.47, -19.23, 54.86]}