Problem 100 - A Graphical Approach to Precalculus with Limits

Question

Incorporate many concepts from earlier work with the method of solving equations involving roots. Work them in order. Consider the equation $$\sqrt[3]{4 x-4}=\sqrt{x+1}$$ Write an explanation of how the solutions of the equation in Exercise 92 relate to the solutions of the original equation. Discuss any extraneous solutions that may be involved.

Step-by-Step Solution

Answer

Solve expanded polynomial for roots; verify to avoid extraneous solutions.

Step 1: Understand the Problem

Our task is to solve the equation \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \). This equation involves a cube root on the left-hand side and a square root on the right-hand side.

Step 2: Eliminate the Cube Root

To eliminate the cube root, we cube both sides of the equation. This gives us:\[(4x - 4) = (\sqrt{x + 1})^3\]Simplifying the right-hand side, we have:\[4x - 4 = (x + 1)^{3/2}\]

Step 3: Eliminate the Square Root

To further simplify, square both sides to eliminate the square root on the right-hand side:\[(4x - 4)^2 = (x + 1)^3\]

Step 4: Expand Both Sides

Calculate the left side first:\[(4x - 4)^2 = 16x^2 - 32x + 16\]Expand the right side:\[(x + 1)^3 = x^3 + 3x^2 + 3x + 1\]

Step 5: Equate and Rearrange

Equating the expanded forms gives:\[16x^2 - 32x + 16 = x^3 + 3x^2 + 3x + 1\]Rearrange all terms to one side of the equation:\[0 = x^3 - 13x^2 + 35x - 15\]

Step 6: Solve the Polynomial Equation

We now solve the polynomial equation \(x^3 - 13x^2 + 35x - 15 = 0\). Depending on previous techniques, use factoring, synthetic division, or numerical methods to find the roots.

Step 7: Check for Extraneous Solutions

Since we squared the equation, extraneous solutions might be introduced. Substitute each solution back into the original equation \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \) to verify validity.Any solution that does not satisfy the original equation is extraneous.

Key Concepts

Cube Roots

Cube roots are the inverse operation of cubing a number. They find what number, when multiplied by itself three times, results in the original number. In our exercise, we encountered the cube root \( \sqrt[3]{4x - 4} \).
Cubing both sides was essential to eliminate the cube root. This action isolates the term we're interested in by removing the cube root from the equation. By cubing both sides, we safely transform the equation into a form that's easier to handle:

• Original equation: \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \)
• After cubing: \( (4x - 4) = ({x + 1})^{3/2} \)

When solving equations with cube roots, always ensure you correctly simplify both sides after cubing. This lets you progress with solving the equation accurately.

Square Roots

Square roots involve finding a number that, when squared, results in the given number. In our problem, the right-hand side involves a square root: \( \sqrt{x + 1} \). Squaring this entire side can help eliminate the square root:

• Original form: \( (x + 1) \)
• Squared form: \( (x + 1)^3 \)
• This leads to: \( (4x - 4)^2 = (x + 1)^3 \)

Squaring was a necessary step to simplify and equate all terms without roots. It's crucial to note that squaring can introduce extra solutions, which we'll discuss later. Always check these solutions derived from a squared equation against the initial equation to determine their validity.

Extraneous Solutions

Extraneous solutions are results that emerge from algebraic manipulations which do not satisfy the original equation. They often arise when squaring or cubing steps are performed. In our exercise, after working through the solution, we needed to check each solution against \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \).
If a solution doesn't satisfy the equation above, it is extraneous. This step is crucial to ensure the solutions are valid for the original problem:

• Solve: Find all possible solutions to the transformed equation.
• Check: Substitute each one back into the original equation.
• Validate: Identify and remove any that don't hold true.

Always remember, when a squared or cubed equation is involved, there is a good chance of encountering extraneous solutions, thus verifying is important.

Polynomial Equations

Polynomial equations are expressions that involve terms with variables raised to natural number powers, like squares or cubes. Our problem eventually converted to a cubic polynomial:
\( x^3 - 13x^2 + 35x - 15 = 0 \).
Polynomials often require specific strategies to solve, such as:

• Factoring: Writing the polynomial as a product of its factors.
• Synthetic Division: An efficient method for dividing polynomials.
• Numerical Methods: When analytical solutions are complex, numerical approximations help.

In this exercise, solving this polynomial allows us to find potential solutions to the transformed equation. Always ensure these are checked back in the original equation to avoid counting any that may be extraneous. Understanding various methods to solve polynomial equations aids immensely in solving complex root-involved problems.