Problem 42 - Algebra and Trigonometry

Question

\(31-42\) . Solve the equation both algebraically and graphically. $$ 6(x+2)^{5}=64 $$

Step-by-Step Solution

Answer

The solution is roughly \(x = -0.4843\), both algebraically and graphically.

Step 1: Simplify the Equation

First, we start by simplifying the given equation algebraically. The equation is given as \(6(x+2)^{5} = 64\). We can divide both sides by 6 to isolate the power term:\[(x+2)^{5} = \frac{64}{6}\]Simplify the fraction:\[(x+2)^{5} = \frac{32}{3}\].

Step 2: Take the Fifth Root

To solve for \(x\), take the fifth root of both sides of the equation:\[x+2 = \left(\frac{32}{3}\right)^{1/5}\].

Step 3: Isolate x

Subtract 2 from both sides to solve for \(x\):\[x = \left(\frac{32}{3}\right)^{1/5} - 2\].

Step 4: Numerically approximate the solution

Using a calculator, compute the value of \(\left(\frac{32}{3}\right)^{1/5}\). This yields approximately 1.5157 (use a calculator for precision). Subtract 2:\[x \approx 1.5157 - 2 = -0.4843\].

Step 5: Set Up the Graph

To solve graphically, define two functions: \(f(x) = 6(x+2)^5\) and \(g(x) = 64\). Plot these functions to find where they intersect.

Step 6: Identify the Intersection Point

On the graph, the intersection point where \(f(x) = g(x)\) occurs at around \(x \approx -0.4843\). This confirms the algebraic solution.

Key Concepts

Algebraic Solution

Solving an equation algebraically involves manipulation and simplification using mathematical operations. For the equation \(6(x+2)^{5} = 64\), our goal is to isolate \(x\). Here’s how it’s done:

First, we divide both sides by 6, simplifying to \((x+2)^{5} = \frac{64}{6}\). This narrows down the problem and simplifies the numbers we're working with.
Next, we simplify \(\frac{64}{6}\) to \(\frac{32}{3}\). The algebraic trick here involves maintaining balance. Whatever we do to one side, we do to the other.

Now, we take the fifth root of both sides: \(x+2 = \left(\frac{32}{3}\right)^{1/5}\). Taking roots is a standard method to undo powers, and here we're tackling the fifth power.
Finally, subtract 2 from both sides to solve for \(x\), giving us \(x = \left(\frac{32}{3}\right)^{1/5} - 2\). This process resolves the equation to its solution form.

Graphical Solution

Graphical methods offer a visual way to solve equations. This involves interpreting the equation in terms of graphs and identifying where solutions occur.
For this example, two functions are defined: \(f(x) = 6(x+2)^5\) and \(g(x) = 64\). The strategy then involves plotting both functions on a graph.

When graphed, the key task is finding where these curves intersect. This point of intersection represents the solution because it's where both functions output the same value.

• Ease of visualization helps confirm calculations.
• It offers a cross-check to algebraic solutions.

While this method requires tools like graphing calculators or software, it provides an intuitive understanding of the problem.

Numerical Approximation

Numerical approximation involves estimating solutions using calculator-aided calculations or iterative methods. It's especially useful when exact algebraic solutions are cumbersome.

Starting from \(x = \left(\frac{32}{3}\right)^{1/5} - 2\), a calculator determines \(\left(\frac{32}{3}\right)^{1/5} \approx 1.5157\). Subtraction yields \(x \approx -0.4843\), providing a precise, usable result.

• Offers a practical and often faster solution route.
• Useful for complex roots or transcendental equations.

This demonstrates an effective blend of algebra with numerical precision, confirming the algebraic finding but without intricate algebraic transformations.

Intersection Point

The intersection point of functions on a graph is crucial in determining solutions for simultaneous or equivalent equations.

In our exercise, the intersection where \(f(x)\) meets \(g(x)\) indicates where the original equation holds true. This point occurs approximately at \(x = -0.4843\). Such points provide not only solutions but also insight into the behavior of equations relative to each other.

• Key for visual confirmation of solutions.
• Useful in understanding function relationships and overlaps.

This intersection offers a visual check against algebraic and numerical solutions, ensuring consistent results across different solving methods.