# y=x^2e^(-x) has a maximum turning point at A(2,4/e^2). The equation x^2e^(-x)-2=0 has 3 real, distinct roots. What are the possible values of k?

## The answers say $0 < k < 2$, but I have no idea how to get to this. Thanks!

Nov 27, 2017

See below.

#### Explanation:

The equation ${x}^{2} {e}^{-} x - 2 = 0$ can be solved with the contribution of the so called Lambert function $W \left(\cdot\right)$

https://en.wikipedia.org/wiki/Lambert_W_function

We have

${x}^{2} {e}^{-} x = 2 \Rightarrow x {e}^{- \frac{x}{2}} = \pm \sqrt{2} \Rightarrow \left(- \frac{x}{2}\right) {e}^{- \frac{x}{2}} = \pm \left(- \frac{\sqrt{2}}{2}\right)$

Now using the property

$X {e}^{X} = Y \Leftrightarrow X = W \left(Y\right)$

we have

$- \frac{x}{2} = W \left(\pm \left(- \frac{\sqrt{2}}{2}\right)\right) \Rightarrow x = - 2 W \left(\pm \left(- \frac{\sqrt{2}}{2}\right)\right)$

with the only real solution

$x = - 2 W \left(\frac{\sqrt{2}}{2}\right) = - 0.901201031729666$

NOTE

Attached the representation of how to obtain a graphical solution as the intersection of ${e}^{-} x$ in blue and $\frac{2}{x} ^ 2$ in red. 