Problem 76 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation involving "nested" radicals for all real solutions analytically. Support your solutions with a graph. $$\sqrt[3]{\sqrt{x+63}}=\sqrt[3]{2 x+6}$$

Step-by-Step Solution

Answer

The only real solution is \(x = 1\).

Step 1: Simplify the Equation

Given the equation, \(\sqrt[3]{\sqrt{x+63}}=\sqrt[3]{2x+6}\). Since both sides are cube roots, we can equate the expressions under the cube roots: \(\sqrt{x+63} = 2x + 6\).

Step 2: Eliminate the Square Root

Square both sides of the equation to eliminate the square root: \((\sqrt{x+63})^2 = (2x + 6)^2\) simplifies to \(x + 63 = 4x^2 + 24x + 36\).

Step 3: Rearrange the Equation

Move all terms to one side to form a quadratic equation: \(0 = 4x^2 + 24x + 36 - x - 63\), which simplifies to \(4x^2 + 23x - 27 = 0\).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For \(4x^2 + 23x - 27\), we have \(a=4\), \(b=23\), \(c=-27\). Calculate the discriminant: \(b^2 - 4ac = 23^2 - 4 \cdot 4 \cdot (-27) = 529 + 432 = 961\). The discriminant is positive, indicating two real solutions.

Step 5: Calculate Roots using Quadratic Formula

Continue with the quadratic formula: \(x = \frac{-23 \pm \sqrt{961}}{8}\). Since \(\sqrt{961} = 31\), the roots are \(x = \frac{-23 + 31}{8}\) and \(x = \frac{-23 - 31}{8}\). Thus, \(x = 1\) and \(x = -\frac{27}{4}\) or \(-6.75\).

Step 6: Verify the Solutions

Substitute \(x = 1\) into the original equation: \(\sqrt[3]{2\cdot1 + 6} = \sqrt[3]{8} = 2\) and \(\sqrt[3]{\sqrt{1+63}} = \sqrt[3]{8} = 2\). Now check \(x = -6.75\): \(\sqrt[3]{2(-6.75) + 6} = \sqrt[3]{-7.5}\) which is not real, so \(x = -6.75\) is not a valid real solution.

Step 7: Graphical Interpretation

Graph both sides: \(y = \sqrt[3]{\sqrt{x+63}}\) and \(y = \sqrt[3]{2x+6}\). Since they intersect at \(x=1\), this confirms the algebraic solution. Solution \(x = 1\) is valid and confirmed through both analytic and graphic methods.

Key Concepts

Solving Equations

Solving equations, especially those involving nested radicals, requires a systematic approach to simplify and find real solutions. In this exercise, we start with an equation that has cube roots on both sides: \[\sqrt[3]{\sqrt{x+63}}=\sqrt[3]{2x+6}\]Here are the essential strategies used to solve this type of equation effectively:

• Equate Inner Expressions: Since the cube roots are the same, equate the expressions inside the roots.
• Eliminate Radicals: To remove the square root, square both sides. Be cautious not to alter the properties of the equation, which could introduce extraneous solutions.
• Reformulate as a Polynomial: This typically involves rearranging terms once radicals are eliminated, converting the problem into a polynomial form.
• Verify Solutions: After finding solutions analytically, always substitute back into the original equation to ensure they fit. Sometimes squaring can introduce solutions that don't actually work in the original equation.

Quadratic Formula

The quadratic formula is a critical tool for solving quadratic equations that arise after simplifying more complex expressions like nested radicals. It provides an exact solution for equations of the form:\[ax^2 + bx + c = 0\]When eliminating radicals leads to a quadratic equation, you can use the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Key steps for applying this formula:

• Identify Coefficients: Ensure you correctly identify the coefficients \(a\), \(b\), and \(c\).
• Calculate the Discriminant: The term \(b^2 - 4ac\) gives insight into the roots. A positive discriminant means two real solutions; zero means one real solution; negative indicates no real solutions.
• Compute the Roots: Solve the equation by substituting your values into the quadratic formula.

In our problem, the quadratic formula helped us find the real solution \(x = 1\), while dismissing \(x = -6.75\) based on verification.

Graphical Solutions

Graphical solutions are an excellent way to visualize the solutions of equations and confirm analytical results. In our exercise, the graphical approach involves plotting both sides of the modified equation:

• Plot \(y = \sqrt[3]{\sqrt{x+63}}\): This function represents the left-hand side.
• Plot \(y = \sqrt[3]{2x+6}\): This represents the right-hand side.
• Find Intersections: The point where these curves intersect represents the solution to the equation. Here, they intersect at \(x = 1\), confirming it's the correct and real solution.

Visual representation helps verify results and provides a deeper understanding of the behavior of the functions involved. It also highlights the practicality and accuracy of algebraic solutions by showing where the calculations meet actual graph outcomes.Graphs were particularly useful in ruling out the extraneous solution \(x = -6.75\), which did not satisfy the real conditions.

Cube Roots

Cube roots, symbolized as \(\sqrt[3]{\cdot}\), are the inverse operations of cubing a number. Understanding how cube roots work is essential for solving problems involving them, like our nested radicals equation:\[\sqrt[3]{\sqrt{x+63}}=\sqrt[3]{2x+6}\]Key insights about cube roots include:

• Real Solutions: Unlike square roots, cube roots always have real solutions, regardless of whether the number inside the root is positive or negative.
• Intersecting with Other Roots: When cube roots are equated with other expressions, understanding their relationship allows you to simplify the problem, as initially seen by equating the inner expressions.
• Behavior in Equations: A function involving cube roots can often appear more linear due to their gradual slope, making them unique compared to other radicals such as squares.

Handling cube roots requires practice, but grasp these basics, and solving related problems will become less daunting. Recognizing their distinguishing characteristics simplifies not only the calculations but also your approach to related tasks.