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.

Also, odd degree polynomials with real coefficients have a real root. $x^3+i$ does not have a real root, for example. — Thomas Andrews, Feb 16 '18 at 14:43
see here https://math.stackexchange.com/questions/7539/finding-all-complex-zeros-of-a-high-degree-polynomial — Dr. Sonnhard Graubner, Feb 16 '18 at 14:58

score 9 · Answer 1 · answered Feb 16 '18 at 14:52

9

$$x^3+7x^2+1=-ixe^{−x}$$ 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.

answered Feb 16 '18 at 14:52

Nathaniel Bubis

33,018

score 1 · Answer 2 · answered Feb 16 '18 at 14:55

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.

score 1 · Answer 3 · answered Feb 16 '18 at 15:01

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.

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

3 Answers3