how many real roots does the equation x^7 + x
have

Question

how many real roots does the equation x^7 + x
have

allank · Answer

Hmm...that is an expression. You need to add "=sumthing" to make it an equation. 
Maybe it's  x^7+x=0 ?

anonymous · Answer

oh sorry i forgot to add that part

allank · Answer

No prob. Let's work it together. 
First we factor out x:   x(x^6+1)=0
okay?

anonymous · Answer

okay

allank · Answer

So that means that either x = 0 (first real root) or x^6+1=0   ok?

anonymous · Answer

i see, so we just solve it by factoring? no need to use the rule of signs or anything fancy?

allank · Answer

We will do something a bit fancy with x^6+1=0
So that means that x^6=-1
Now, there is no real value for x that can be raised to power 6, to get a value of -1. 
That means that there is only one real root for the original equation: x=0

anonymous · Answer

okay, thanks for the help

allank · Answer

You're welcome :)