The equation $x^3+7x^2+1+ixe^{-x}=0$ has no real solution.

Question

The equation $x^3+7x^2+1+ixe^{-x}=0$(where $i=\sqrt(-1)$) has

1) no real solution.

2) exactly one real solution.

3) exactly two real solution.

4) exactly three real solution.

We know that odd degree polynomial at least one real zeros but the above equation is not polynomial. Please help. Thanks.

Answer 1

$$x^3+7x^2+1=-ixe^{−x}$$ Clearly, if $x\neq0$ is real, the RHS is imaginary, while the LHS is real, a contradiction. $x$ cannot be zero by simple observation, so that we can safely say there are no real solutions.

Answer 2

If this equaiton had a real solution $x \in \mathbb{R}$, then $$x^3 + 7x^2 + 1 +ix e^{-x} = 0$$ and $x^3 + 7x^2 + 1= 0$ and $x e^{-x}=0$. From the second equation, we see that $x=0$, but this doesn't verify the first one, so the equation $$x^3 + 7x^2 + 1 +ix e^{-x} = 0$$ has no real solution.

Answer 3

If $x\in\mathbb{R}$ satisfies $x^3+7x^2+1+ixe^{-x}=0$, then

$\left\{ \begin{array}{} x^3+7x^2+1=0\\ xe^{-x}=0 \end{array}\right. $

The second equality implies that $x=0$, but $0$ is not a solution for the first equality. Therefore the equation has no real solution.