How do you solve #x^5+9x>=10x^3#?

1 Answer
Dec 15, 2016

Answer:

The answer is #x in [-3,-1] uu [0, 1] uu [3,+ oo [ #

Explanation:

Let #f(x)=x^5-10x^3+9x#

Let's rearrange the equation

#x^5-10x^3+9x>=0#

#x(x^4-10x^2+9)>=0#

#x(x^2-1)(x^2-9)>=0#

#x(x+1)(x-1)(x+3)(x-3)>=0#

We can do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##-1##color(white)(aaaa)##0##color(white)(aaaa)##1##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(aaa)##+##color(white)(aaa)##+#

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

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaa)##-##color(white)(aaa)##+##color(white)(aaa)##+#

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

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

So,

#f(x)>=0# when #x in [-3,-1] uu [0, 1] uu [3,+ oo [ #