Problem 131
Question
Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$2 x^{2} e^{2 x}+2 x e^{2 x}=0$$
Step-by-Step Solution
Verified Answer
The solutions to the equation \(2x^2e^{2x} + 2xe^{2x}=0\) are \(x=0\) and \(x=-1\), rounded to three decimal places.
1Step 1: Factor the Equation
The given equation, \(2x^2e^{2x} + 2xe^{2x} = 0\), can be factored such that common terms are grouped together. Factoring gives: \(2xe^{2x}(x+1) = 0\).
2Step 2: Solve Equation for Roots
To solve for \(x\), the equation \(2xe^{2x}(x+1) = 0\) has two possible solutions. These are computed by setting each factor equal to zero: Solution 1: \(2xe^{2x} = 0\), which gives \(x = 0\). Solution 2: \(x+1 = 0\), which gives \(x = -1\).
3Step 3: Round Off Solutions
The solutions found are: \(x = 0\) and \(x = -1\). No need for rounding off because they are already exact to three decimal places.
4Step 4: Verify Using a Graphing Utility
To verify the solutions graphically, use a graphing utility to plot the function \(y=2x^2e^{2x}+2xe^{2x}\) and check where the curve intersects the x-axis. The function becomes zero at the points of intersection, which should match our computed solutions of \(x = 0\) and \(x = -1\).
Key Concepts
Exponential FunctionFactoring EquationsGraphical Solution Verification
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the equation given, the term \(e^{2x}\) is an exponential component. Here's what you should know about exponential functions:
- They involve the constant \(e\), approximately equal to 2.71828, which is the base of natural logarithms.
- The variable in the exponent makes these functions grow or decay at increasingly rapid rates, appearing in many natural processes like compound interest or population growth.
- In our equation \(2x^2e^{2x} + 2xe^{2x} = 0\), the exponential factor \(e^{2x}\) influences both terms.
- The presence of the exponential function often necessitates factoring or other algebraic methods to solve when combined with polynomial terms, as seen here.
Factoring Equations
Factoring is a powerful algebraic technique used to simplify an equation and identify its roots. When you factor an equation, you break it down into simpler terms known as 'factors'. For the given equation, factoring was a key step:
- The original equation \(2x^2e^{2x} + 2xe^{2x} = 0\) contains common elements like \(2xe^{2x}\) in both terms.
- By extracting the common factor \(2xe^{2x}\), you simplify the expression to \(2xe^{2x}(x+1) = 0\).
- Once factored, you use the Zero Product Property. This states that if a product of several factors equals zero, then at least one of the factors must be zero.
- Here, it leads to two solutions: \(x = 0\) from \(2xe^{2x} = 0\) (keeping in mind that \(e^{2x}\) never equals zero), and \(x = -1\) from \(x+1 = 0\).
Graphical Solution Verification
Once you've solved an equation algebraically, verifying the solutions graphically is a great way to ensure accuracy. With graphing utilities or tools, you can visualize the function and its intersections with the x-axis.
- By plotting the function \(y = 2x^2e^{2x} + 2xe^{2x}\), you look for points where the graph crosses the x-axis.
- These points of intersection correspond to the x-values where the function is zero, which confirms the algebraic solutions.
- In the step-by-step solution, the computed solutions were \(x = 0\) and \(x = -1\). When graphing, these should appear as intersections with the x-axis on the plot.
- This visual confirmation helps confirm that the calculations were correctly done and increases confidence in the solution.
Other exercises in this chapter
Problem 130
Use a graphing utility to approximate the point of intersection of the graphs. Round your result to three decimal places. $$\begin{aligned}&y_{1}=1.05\\\&y_{2}=
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