Thanks Geno,

I am just going to work through my version of your logic.  :)

$$\\x^{15}+x^5+2=0\\\\
Let\;\;y=x^5\\\\
y^3+y+2=0\\\\
y(y^2+1)=-2\\\\
y^2+1 $ must be positive so $y$ must be negative. Hence$\\\\
x^5$ must be negative, hence $\\\\
x $ must be negative$\\\\
$Therefore, no positive roots exist.$$$

prove that x^15+x^5+2=0, do not have positive roots?

 

Let's write x^5 = y

Then we get:

y^3 +y +2 = 0

 

y^3 +y = -2

 

y(1 +y^2) = -2 

 

y =-2 or y^2 +1 =-2

 

 

We solve y^2 +1 =-2

 

y^2 = -3

 

y doesn't have any solution among the real numbers R.

 

y = √-3 = √3 *i or -√3 *i

 

To sum it up: y = -2 or y = √3 *i or y = -√3 *i

x^5 = y

 

...

Let's write x^5 = y     good idea

Then we get:

y^3 +y +2 = 0

 y^3 +y = -2

 y(1 +y^2) = -2       Yes

 y =-2 or y^2 +1 =-2       WHY?

You are right, Melody!

 

I must have been asleep :P

 

 

Ok, I'll try once more:

 

 

Let's write x^5 = y

Then we get:

y^3 +y +2 = 0

 

y^3 +y = -2

 

y(1 +y^2) = -2 

 

Ok, we know that:

y^2 +1 = (y- i)(y+i)

 

I'm not really sure from here :(

I would argue that if you want the equation to have a positive root, then x must be positive.

But if x is positive, then both x^15 is positive and x^5 is positive.

Adding two positive number to the number 2, your answer must be positive; thus, it can't be zero.

(proof by contradiction)