Problem 36 - Algebra and Trigonometry

Question

Solve the equation. \(X^{2} e^{X}+X e^{X}-e^{X}=0\)

Step-by-Step Solution

Answer

The solutions are \( X = \frac{-1 + \sqrt{5}}{2} \) and \( X = \frac{-1 - \sqrt{5}}{2} \).

Step 1: Factor the Common Term

First, recognize that each term in the equation contains a factor of \( e^X \). Factor \( e^X \) out of the expression:\[ e^X (X^2 + X - 1) = 0 \]

Step 2: Solve for \( e^X = 0 \)

Since \( e^X \) is never zero for any real number \( X \), the equation \( e^X = 0 \) does not contribute any solutions.

Step 3: Solve the Quadratic Equation

Now solve the quadratic equation \( X^2 + X - 1 = 0 \) using the quadratic formula:\[ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \). Substitute these into the formula:\[ X = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]Simplify the expression:\[ X = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

Key Concepts

Factoring

One important skill in solving complex equations is recognizing and using common factors. In the equation \( X^2 e^{X} + X e^{X} - e^{X} = 0 \), each term includes the factor \( e^X \). By factoring out \( e^X \) from the equation, the expression becomes simpler:

• \( e^X (X^2 + X - 1) = 0 \)

Factoring helps in breaking down complex expressions into more manageable parts. This step reveals two separate components: the exponential part \( e^X \) and the quadratic expression \( X^2 + X - 1 \).
The next step is to solve each component separately. However, in this equation, we note that \( e^X \) never equals zero, so we focus on the quadratic part \( X^2 + X - 1 \).
Factoring becomes a strategic tool in mathematical problem-solving, enabling simplification and solving of various kinds of equations by isolating difficult parts. Remember that finding and pulling out common factors is crucial for simplifying and solving complex equations.

Quadratic Formula

Quadratic equations are standard in algebra, and they come in the form \( ax^2 + bx + c = 0 \). To solve these equations, the quadratic formula is a powerful tool, expressed as:

• \( X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula can find solutions for any quadratic equation, as long as you correctly identify \( a \), \( b \), and \( c \), which are the coefficients in the equation.
In the equation \( X^2 + X - 1 = 0 \), these coefficients are \( a = 1 \), \( b = 1 \), and \( c = -1 \).
By substituting these into the formula, we get:

• \( X = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \)
• \( X = \frac{-1 \pm \sqrt{5}}{2} \)

This results in two possible solutions for \( X \), since the formula accounts for the "\( \pm \)" that offers both a positive and negative solution.
Understanding the quadratic formula helps solve a wide range of problems that involve quadratic equations, making it an essential concept in algebra.

Exponential Functions

Exponential functions appear in various mathematical contexts and often take the form \( a \cdot e^{x} \), where \( e \) is a constant approximately equal to 2.71828. These functions grow or decay at exponential rates and are characterized by their rapid increase or decrease.
In the equation \( X^2 e^X + X e^X - e^X = 0 \), the exponential part \( e^X \) doesn't contribute any solutions since \( e^X \) is never zero for any real value of \( X \).
It's crucial to recognize when an exponential term can be factored out and when it might equal a specific value. In this case, because the exponential function is always positive and non-zero, solving

• \( e^X = 0 \)

would not work. Instead, focus shifts only to the remaining polynomial part.
Understanding and working with exponential functions plays a vital role in calculus and beyond, particularly in representing growth and decay in practical and theoretical models.