Solving The Cube Root Equation $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$

In the realm of algebra, solving equations involving radicals, especially cube roots, is a fundamental skill. This article aims to provide a comprehensive guide on how to solve equations of the form $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$. We will dissect the problem, explore the correct methodology, and delve into the underlying mathematical principles. Understanding these concepts is crucial for students, educators, and anyone interested in mathematics.

Understanding the Problem

The given equation is $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$. This equation involves cube roots, which are different from square roots in that they can handle negative numbers. The presence of cube roots on both sides suggests a specific approach to solving this equation. Before diving into the solution, it's essential to understand the properties of cube roots and how they interact with algebraic operations. Cube roots are the inverse operation of cubing, meaning that if you cube a cube root, you get the original number back. This property is the key to solving equations involving cube roots.

In essence, solving this equation requires us to isolate the variable x. The challenge lies in the presence of the cube root, which obscures the underlying polynomial equation. Our primary strategy will be to eliminate the cube root by applying the inverse operation, which is cubing. Once we remove the cube roots, we will be left with a more manageable equation, likely a polynomial equation, which we can then solve using standard algebraic techniques. This process may involve factoring, using the quadratic formula, or other methods depending on the degree of the polynomial. The critical first step, however, is to eliminate the cube roots, which will transform the equation into a more familiar form.

Identifying the Correct Solution Strategy

When faced with the equation $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$, the initial question is how to eliminate the cube roots. Let's analyze the options provided:

A. Square both sides and then solve the resulting quadratic equation. B. Square both sides and then solve the resulting cubic equation. C. Cube both sides and then solve the resulting equation.

Option A and B suggest squaring both sides. However, squaring both sides would still leave us with radicals, albeit of a different form. Squaring a cube root doesn't eliminate the radical; it results in a power of 2/3. This approach would complicate the equation further, making it harder to solve. Therefore, squaring both sides is not the correct first step in this scenario. It's crucial to choose an operation that directly cancels out the cube root, and squaring does not achieve this.

Option C, which involves cubing both sides, is the correct approach. Cubing is the inverse operation of taking the cube root. When we cube a cube root, the radical is eliminated, leaving us with the expression inside the radical. This simplifies the equation and allows us to work with a polynomial equation, which is much easier to solve. Cubing both sides will transform the equation from one involving radicals to a standard algebraic equation, which can then be solved using familiar methods. Therefore, cubing both sides is the key to unlocking the solution to this equation.

Step-by-Step Solution: Cubing Both Sides

The correct approach to solving $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$ is to cube both sides of the equation. This eliminates the cube root, simplifying the equation into a more manageable form. Let's perform this step:

$(\sqrt[3]{x^2-6})^3=(\sqrt[3]{2x+2})^3$

When we cube a cube root, the cube root and the cube cancel each other out. This is because cubing is the inverse operation of taking the cube root. Applying this principle to both sides of the equation, we get:

$x^2-6 = 2x+2$

Now, we have a quadratic equation. Quadratic equations are polynomial equations of the second degree and can be written in the general form $ax^2 + bx + c = 0$, where a, b, and c are constants, and a is not equal to zero. Solving a quadratic equation typically involves rearranging it into the standard form and then using methods such as factoring, completing the square, or applying the quadratic formula. The goal is to find the values of x that satisfy the equation, which are also known as the roots or solutions of the equation. Transforming the original equation into a quadratic equation is a significant step forward in finding the solution.

Solving the Resulting Equation

After cubing both sides of the original equation, we arrived at the quadratic equation $x^2 - 6 = 2x + 2$. The next step is to solve this quadratic equation. To do this, we first need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. This involves moving all terms to one side of the equation, leaving zero on the other side.

Subtracting $2x$ and $2$ from both sides, we get:

$x^2 - 2x - 8 = 0$

Now we have a quadratic equation in the standard form. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, the equation can be easily factored. Factoring involves expressing the quadratic expression as a product of two binomials. We are looking for two numbers that multiply to -8 and add up to -2. These numbers are -4 and 2. Therefore, we can factor the quadratic equation as follows:

$(x - 4)(x + 2) = 0$

To find the solutions, we set each factor equal to zero:

$x - 4 = 0$ or $x + 2 = 0$

Solving these linear equations gives us the solutions:

$x = 4$ or $x = -2$

Thus, the solutions to the quadratic equation are $x = 4$ and $x = -2$. These are the potential solutions to the original cube root equation. However, it is crucial to verify these solutions in the original equation to ensure they are not extraneous solutions.

Verifying the Solutions

After solving the quadratic equation, we obtained two potential solutions: $x = 4$ and $x = -2$. However, it is essential to verify these solutions in the original equation, $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$, to ensure they are not extraneous solutions. Extraneous solutions are solutions that arise during the solving process but do not satisfy the original equation. This can happen when we perform operations that are not reversible, such as squaring or, in this case, cubing both sides of an equation.

Let's first check $x = 4$:

$\sqrt[3]{(4)^2-6}=\sqrt[3]{2(4)+2}$

$\sqrt[3]{16-6}=\sqrt[3]{8+2}$

$\sqrt[3]{10}=\sqrt[3]{10}$

The equation holds true for $x = 4$, so it is a valid solution.

Now, let's check $x = -2$:

$\sqrt[3]{(-2)^2-6}=\sqrt[3]{2(-2)+2}$

$\sqrt[3]{4-6}=\sqrt[3]{-4+2}$

$\sqrt[3]{-2}=\sqrt[3]{-2}$

This equation also holds true for $x = -2$, so it is also a valid solution.

Therefore, both $x = 4$ and $x = -2$ are solutions to the original equation. Verifying solutions is a critical step in solving radical equations, as it ensures that the solutions we obtain are genuine and not artifacts of the solving process. This step adds rigor to the solution and confirms that our answers are correct.

Final Answer

In summary, to solve the equation $\sqrt[3]{x^2-6}=\sqrt[3]{2x+2}$, the correct approach is to:

1. Cube both sides of the equation to eliminate the cube roots.
2. Solve the resulting quadratic equation.
3. Verify the solutions in the original equation to ensure they are not extraneous.

By following these steps, we transformed the cube root equation into a quadratic equation, solved it, and verified the solutions. This process demonstrates a fundamental technique in algebra for dealing with radical equations. The solutions to the equation are $x = 4$ and $x = -2$, both of which satisfy the original equation. Therefore, we have successfully solved the problem and provided a comprehensive explanation of the solution process.

This detailed explanation covers the initial setup, the step-by-step solution, and the verification process, providing a thorough understanding of how to solve cube root equations. Understanding these techniques is crucial for success in algebra and beyond.