Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation by hand. Do not use a calculator. $$\sqrt[3]{2 x^{2}+1}=\sqrt[3]{1-x}$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = -\frac{1}{2}\).

Step 1: Equalize the Cube Roots

The given equation is \( \sqrt[3]{2x^2 + 1} = \sqrt[3]{1 - x} \). Since both sides are cube roots, we can remove the cube roots by equating the radicands: \(2x^2 + 1 = 1 - x \).

Step 2: Rearrange the Equation

Rearrange the equation to bring all terms to one side. This gives \(2x^2 + x + 1 - 1 = 0\), which simplifies to \(2x^2 + x = 0\).

Step 3: Factor the Equation

Factor out the common factor \(x\) from the equation \(2x^2 + x = 0\). This gives \(x(2x + 1) = 0\).

Step 4: Solve for x

Set each factor equal to zero and solve for \(x\):1. \(x = 0\)2. \(2x + 1 = 0\), solving for \(x\) gives \(x = -\frac{1}{2}\).

Step 5: Verify the Solutions

Substitute \(x = 0\) and \(x = -\frac{1}{2}\) back into the original equation to verify:For \(x = 0\):\(\sqrt[3]{2(0)^2 + 1} = \sqrt[3]{1} = 1\) and \(\sqrt[3]{1 - 0} = \sqrt[3]{1} = 1\).For \(x = -\frac{1}{2}\):\(\sqrt[3]{2(-\frac{1}{2})^2 + 1} = \sqrt[3]{1.5}\) and \(\sqrt[3]{1 - (-\frac{1}{2})} = \sqrt[3]{1.5}\). Both substitutions verify the original equation.

Key Concepts

Cube Root Equations

Cube root equations involve variables under a cube root sign. These equations can often be solved by eliminating the cube root. To do this, you raise both sides of the equation to the power of three, which removes the cube roots. For example, if you have an equation \(\sqrt[3]{2x^2 + 1} = \sqrt[3]{1 - x}\), you can simply equate the insides, or the radicands:

• Set \(2x^2 + 1 = 1 - x\).

By solving the equation without the cube roots, you simplify the process significantly.By getting rid of cube roots, you convert the problem into a polynomial equation, which is often much simpler to handle. Remember to check your solutions at the end to make sure they hold true in the original equation, as sometimes the process can introduce extraneous solutions.

Factoring Polynomials

Factoring is like finding numbers that multiply together to get another number. In polynomials, it's finding expressions that multiply to become the original polynomial. Consider this equation:\(2x^2 + x = 0\).The process involves looking for common factors first. Here, \(x\) is a common factor:

• Factor out \(x\), leading to \(x(2x + 1) = 0\).
• Set each part of the factored equation to zero: \(x = 0\) and \(2x + 1 = 0\).
• Solving the second equation gives \(x = -\frac{1}{2}\).

Factoring is powerful because it transforms complicated equations into simpler ones, where solutions are easier to find. Make sure to try factoring first with zero as the goal when you see a polynomial equation. Factoring helps isolate variables, a key step in solving many equations.

Verification of Solutions

Verification is the final and crucial step in solving equations. Its purpose is to confirm that the solutions obtained actually satisfy the original equation. This is especially important if we've altered the equation, such as removing cube roots or factoring.For our equation \(\sqrt[3]{2x^2 + 1} = \sqrt[3]{1 - x}\), we found \(x = 0\) and \(x = -\frac{1}{2}\) as possible solutions. Let's verify:

• Substitute \(x = 0\): - Left: \(\sqrt[3]{2(0)^2 + 1} = \sqrt[3]{1} = 1\). - Right: \(\sqrt[3]{1 - 0} = \sqrt[3]{1} = 1\).
• Substitute \(x = -\frac{1}{2}\): - Left: \(\sqrt[3]{2(-\frac{1}{2})^2 + 1} = \sqrt[3]{1.5}\). - Right: \(\sqrt[3]{1 - (-\frac{1}{2})} = \sqrt[3]{1.5}\).

Both substitutions verify the original equation is satisfied by these solutions.By verifying, you ensure that every solution you find genuinely works, providing confidence in your problem solving. It also helps catch any mistakes made along the way. Always take this step to ensure accuracy.