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how many real roots does the equation x^7 + x have
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Hmm...that is an expression. You need to add "=sumthing" to make it an equation. Maybe it's x^7+x=0 ?
oh sorry i forgot to add that part
No prob. Let's work it together. First we factor out x: x(x^6+1)=0 okay?
okay
So that means that either x = 0 (first real root) or x^6+1=0 ok?
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i see, so we just solve it by factoring? no need to use the rule of signs or anything fancy?
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
okay, thanks for the help
You're welcome :)
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