Problem Analysis

The given equation is:

$x^2 - ke^x = 0$

The problem asks us to determine the range of $k$ such that the equation has two distinct real roots.

Step-by-Step Solution

1. Rearranging the Equation

Rewriting the equation:

$x^2 = ke^x$

This represents a relationship between $x$ and $k$. For the equation to have two distinct real roots, the two curves $y = x^2$ (a parabola) and $y = ke^x$ (an exponential function scaled by $k$) must intersect exactly at two points.

2. Properties of Intersection

The number of intersections depends on the behavior of the two curves:

• $y = x^2$: A parabola symmetric about the $y$-axis.
• $y = ke^x$: An exponential curve that increases for $x > 0$ and approaches $0$ as $x \to -\infty$.

The key is to analyze the conditions for the two curves to intersect at two distinct points:

1. The exponential curve must "rise quickly enough" relative to the parabola for there to be two intersections.
2. This means $k$ must control the steepness of the exponential curve.

3. Differentiating to Find Turning Points

To analyze the intersections, consider the function:

$f(x) = x^2 - ke^x$

The roots of this function correspond to the intersections. To ensure two distinct roots, we investigate its critical points:

$f'(x) = 2x - ke^x$

Setting $f'(x) = 0$:

$2x = ke^x \quad \implies \quad k = \frac{2x}{e^x}$

Define $g(x) = \frac{2x}{e^x}$, which determines the values of $k$ where $f'(x) = 0$.

4. Behavior of $g(x)$

The function $g(x) = \frac{2x}{e^x}$:

1. Is defined for $x > 0$.
2. Has a maximum at some $x = x_0 > 0$, where $g'(x) = 0$.

Differentiating $g(x)$:

$g'(x) = \frac{(2)(e^x) - (2x)(e^x)}{(e^x)^2} = \frac{2(e^x - x \cdot e^x)}{e^{2x}}$

$g'(x) = \frac{2e^x(1 - x)}{e^{2x}} = \frac{2(1 - x)}{e^x}$

Setting $g'(x) = 0$:

$1 - x = 0 \quad \implies \quad x = 1$

Thus, $g(x)$ has a maximum at $x = 1$. Substituting $x = 1$ into $g(x)$:

$g(1) = \frac{2(1)}{e^1} = \frac{2}{e}$

5. Range of $k$

For the function $f(x)$ to have two distinct roots:

1. The value of $k$ must be less than the maximum value of $g(x)$, i.e., $k < \frac{2}{e}$.
2. Additionally, $k > 0$ because $k$ must scale the exponential function.

Thus, the range of $k$ is:

$0 < k < \frac{2}{e}$

Final Answer

The range of $k$ for which the equation $x^2 - ke^x = 0$ has two distinct real roots is:

$\boxed{0 < k < \frac{2}{e}}$

Do you have any questions about the solution or would you like further clarification? Below are five related questions for further exploration.

Related Questions

1. What happens to the number of roots when $k = \frac{2}{e}$?
2. How does the intersection behavior change if $k > \frac{2}{e}$?
3. Can the equation have exactly one real root? If so, for which value of $k$?
4. How does the symmetry of $y = x^2$ affect the root distribution for negative $k$?
5. What are the implications of this result for practical applications involving exponential and quadratic functions?

Tip

When analyzing intersections of curves, consider plotting both functions to visually verify the number of solutions. This can provide intuition for critical points and parameter ranges.