Problem 61 - A Graphical Approach to College Algebra

Question

Incorporate many concepts from Chapter 3 with the method of solving equations involving roots. Work them in order. Consider the equation $$\sqrt[3]{4 x-4}=\sqrt{x+1}$$ Use both an analytic method and your calculator to solve the original equation.

Step-by-Step Solution

Answer

The solutions of the equation are \( x \approx 2.55 \) and \( x \approx 0.45 \).

Step 1: Identify the Equation

Given the equation \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \) which involves roots on both sides.

Step 2: Cube Both Sides

To eliminate the cube root, raise both sides of the equation to the power of 3: \( (\sqrt[3]{4x - 4})^3 = (\sqrt{x + 1})^3 \). This simplifies to \( 4x - 4 = (\sqrt{x + 1})^3 \).

Step 3: Square and Rearrange

Now, to eliminate the remaining square root, square both sides: \( (4x - 4)^2 = x + 1 \). Simplify further to formulate into a single equation: \( 16x^2 - 32x + 16 = x + 1 \).

Step 4: Simplify and Solve the Quadratic Equation

Rearrange the equation to standard quadratic form: \( 16x^2 - 33x + 15 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 16 \), \( b = -33 \), \( c = 15 \).

Step 5: Calculate the Discriminant

Find the discriminant: \( b^2 - 4ac = (-33)^2 - 4(16)(15) = 1089 - 960 = 129 \). Since the discriminant is positive, there are two real solutions.

Step 6: Find the Solutions

Calculate the roots using the quadratic formula: \( x = \frac{33 \pm \sqrt{129}}{32} \). Compute these values: \( x = \frac{33 + \sqrt{129}}{32} \) and \( x = \frac{33 - \sqrt{129}}{32} \).

Step 7: Verify the Solutions

Verify the solutions by substituting them back into the original equation \( \sqrt[3]{4x - 4} = \sqrt{x + 1} \) to ensure they satisfy it.

Step 8: Use a Calculator for Verification

Use a calculator to solve the original equation and verify the analytical solutions by checking if \( 4x-4 = (\sqrt{x+1})^3 \) holds true with the calculated roots.

Key Concepts

Analytic Method

The analytic method is a mathematical approach used to find precise solutions to equations by manipulating them through algebraic operations. When you solve an equation analytically, you aim to isolate the unknown variable by performing transformations such as expanding, factoring, or rearranging the terms. This method is especially useful for finding exact solutions, which is critical when dealing with intricate functions involving roots or exponents.
In the provided exercise, the analytic method begins by simplifying the equation \(\sqrt[3]{4x - 4} = \sqrt{x + 1}\). This initial setup reflects a challenge due to both cube and square roots. Using the analytic method allows us to systematically address each part by first eliminating the cube root, then the square root, to isolate and untangle the variable \(x\). Here's a breakdown of the process:

• Cube both sides to remove the cube root.
• Then square the remaining expression to eliminate the square root.
• Rearrange the resulting expression into a standard quadratic form for further analysis.

This process highlights the systematic approach of the analytic method, showing its precision in handling complex expressions.

Roots of Equations

Roots of equations are the values of the variable that satisfy the equation, meaning they make the equation true when substituted into it. In the context of polynomial equations like quadratic ones, roots may be real or complex numbers, depending on the nature of the equation and its discriminant.
After converting the original equation into a quadratic form \(16x^2 - 33x + 15 = 0\), our goal becomes to discover its roots. Calculating these roots involves several key steps:

• Ensure the equation is in the standard form \(ax^2 + bx + c = 0\).
• Calculate the discriminant \(b^2 - 4ac\) to determine the nature of the roots. A positive discriminant indicates two distinct real roots.
• Apply the quadratic formula to find the precise value for each root, crucial for accurately solving the original problem.

Finding the roots helps in verifying if they satisfy the original equation, thus confirming the solution.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This formula provides a straightforward way of calculating the roots of the equation. It is derived from the process of completing the square and can solve any quadratic equation, whether the roots are real or complex.
The formula is written as:

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

In our exercise, the quadratic formula is employed to solve the rearranged equation \(16x^2 - 33x + 15 = 0\). By plugging the coefficients \(a = 16\), \(b = -33\), and \(c = 15\) into the formula, we calculate the roots. To complete this:

• Find the discriminant to check it is positive, which guarantees real solutions.
• Use the formula to compute the roots, offering us \(x = \frac{33 \pm \sqrt{129}}{32}\).

The quadratic formula's universal applicability makes it indispensable for solving quadratic equations, confirming the validity of the analytic solutions through precise calculation.