How do you solve #((x+7)(x-3))/(x-5)^2>0#?

1 Answer
Dec 20, 2016

Answer:

The answer is #x in ] -oo,-7 [ uu ] 3,5 [uu ] 5,+oo [ #

Explanation:

Let #f(x)=((x+7)(x-3))/(x-5)^2#

The domain of #f(x)# is #D_f(x)=RR-{5}#

#AA x in D_f(x)#, the denominator is #>0#

We can now do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-7##color(white)(aaaa)##3##color(white)(aaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+7##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)()##∥##color(white)(aa)##+#

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

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

Therefore,

#f(x)>0# when #x in ] -oo,-7 [ uu ] 3,5 [uu ] 5,+oo [ #

graph{((x+7)(x-3))/(x-5)^2 [-23.16, 34.55, -11.88, 16.99]}