1 answer

How do you solve #(x^2+5x)/(x-3)>=0# using a sign chart?

Question:

How do you solve #(x^2+5x)/(x-3)>=0# using a sign chart?

Answers

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

Explanation:

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 the sign chart

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

#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##+##color(white)(aaaa)##color(red)(∥)##color(white)(aa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##+##color(white)(aaaa)##color(red)(∥)##color(white)(aa)##+#

#color(white)(aaaa)##x-3##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaa)##color(red)(∥)##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##-##color(white)(aaaa)##color(red)(∥)##color(white)(aa)##+#

Therefore,

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

.

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