This ain't a Joke!

This is the non-Calculus solution :

At the get-go, factor out the $x$ from the equation. Then,

$\displaystyle \large {\wp(x)} = x(3x^{4} -25x^{2} +60)$

This gives the trivial root of $x = 0$ .

Set $\displaystyle t = x^2$ . Then, the bi-quadratic reduces to :

$\displaystyle 3t^{2} -25t +60$

By eyeballing it, one can determine that the $\large D$ for this quadratic is less than 0, consequently proving that the said quadratic( and hence bi-quadratic) has no real solutions.

Hence, the number of real roots of $\displaystyle \large {\wp(x)}$ , $\displaystyle \large N = \boxed{1}$

graph shows that it has only one real solution at $x= 0$ , and we can also prove it like this.

$\displaystyle \large {\wp(x)} = x(3x^4 - 25x^2 + 60)$

This gives the trivial root at $x=0$

now let $x^2 = K$

new equation comes as $3k^2 -25k + 60$

since its determinant is negetive so it will not provide any real solution.

Hence, the number of real roots of , $\displaystyle \large {\wp(x)}$ , $= 1$

$\wp(x)=x(3x^4-25x^2+60)$

Let $x^2=y \Rightarrow x(3y^2-25y+60)$

Solving $3y^2-25y+60$ we get that $y=\frac{25 \pm \sqrt{-95}}{6} \therefore \boxed{x=0}$ in the only real root.

Real roots come in pairs. Th only possible values are 5,3,1. Type them in and you get the right answer!