Problem 36 - Precalculus Mathematics for Calculus

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 out the common term

Begin by observing that each term in the equation \(x^2 e^x + x e^x - e^x = 0\) contains \(e^x\). Factor \(e^x\) out, resulting in: \[ e^x (x^2 + x - 1) = 0 \]

Step 2: Solve the exponential equation

The equation \(e^x (x^2 + x - 1) = 0\) implies that either \(e^x = 0\) or \(x^2 + x - 1 = 0\). However, \(e^x\) is never zero for any real number \(x\). So we move to the next component.

Step 3: Solve the quadratic equation

Focus on solving \(x^2 + x - 1 = 0\). This can be solved using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -1\).

Step 4: Apply the quadratic formula

Calculate the discriminant \(b^2 - 4ac\): \(1^2 - 4(1)(-1) = 1 + 4 = 5\). Now, substituting into the quadratic formula gives: \[ x = \frac{-1 \pm \sqrt{5}}{2} \]

Step 5: Simplify the roots

The two solutions to the quadratic equation are \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\).

Key Concepts

Exponential Functions

Exponential functions feature prominently in mathematics and many real-world applications. The exponential function is particularly useful because it describes growth and decay processes, among others. In its basic form, an exponential function can be written as \( e^x \), where \( e \) (approximately 2.71828) is the base of the natural logarithm. When graphed, \( e^x \) displays a distinct curve that rises sharply as \( x \) increases.In our exercise, we encounter an exponential function \( e^x \), which plays a crucial role within the given equation \( x^2 e^x + x e^x - e^x = 0 \). Knowing that \( e^x \) is never zero for any real number \( x \) is important. This helps us realize why the product \( e^x(x^2 + x - 1) = 0 \) reduces to solving the condition \( x^2 + x - 1 = 0 \) instead.

Factoring

Factoring is an essential algebraic strategy used to simplify equations or expressions and make them more manageable. The process involves breaking down an expression into products of simpler factors. Factoring is particularly useful when dealing with polynomial equations or expressions like in our current exercise.In the equation \( x^2 e^x + x e^x - e^x = 0 \), factoring helps streamline the equation. Each term includes the common factor \( e^x \). By factoring \( e^x \) out, we simplify our expression to \( e^x(x^2 + x - 1) = 0 \). This simplification is vital for identifying the conditions under which the equation is satisfied, allowing us to focus on the quadratic equation \( x^2 + x - 1 = 0 \) without the complication of the exponential term.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). It provides a straightforward way to find the solutions, or roots, of these equations. The quadratic formula is given as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By plugging in the coefficients \( a \), \( b \), and \( c \), one can determine the roots of the quadratic equation. In our example, the expression \( x^2 + x - 1 = 0 \) is addressed using this formula. With \( a = 1 \), \( b = 1 \), and \( c = -1 \), we calculate the discriminant as \( b^2 - 4ac = 5 \). This leads to two possible solutions achieved by substituting into the formula, giving us \( x = \frac{-1 \pm \sqrt{5}}{2} \). This result illustrates how quadratic equations can have two real, and in this case, irrational roots.

Roots of Equations

Finding the roots of an equation, or solving for when an equation equals zero, is a fundamental endeavor in mathematics. These roots are the values of the variable that make the equation true. In the context of our problem, solving \( e^x(x^2 + x - 1) = 0 \) after factoring reveals potential roots.Although \( e^x \) itself has no real roots, since it is never zero, the equation \( x^2 + x - 1 = 0 \) does. Applying the quadratic formula, we identified the roots as \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \). These solutions represent the points where our reduced polynomial equation equals zero, fulfilling the initial equation's requirements. Roots like these often have significant analytical importance and occasional graphical significance in depicting where functions intersect the axis on a graph.