Solving The Equation: $(x^2-8x+20)^{1/3} = 2$

Introduction

In this article, we'll dive deep into solving the equation $(x^2-8x+20)^{\frac{1}{3}}=2$. This is a fascinating problem that combines algebraic manipulation with a touch of equation-solving techniques. Whether you're a student tackling math homework or just a math enthusiast, you'll find this step-by-step guide helpful. We will break down each step to make it clear and easy to follow. So, let's get started and unravel this mathematical puzzle together!

Understanding the Equation

Before we jump into solving, let's understand the equation $(x^2-8x+20)^{\frac{1}{3}}=2$. This equation involves a fractional exponent, which can sometimes seem intimidating, but don't worry, we'll break it down. The expression $(x^2-8x+20)^{\frac{1}{3}}$ means we are taking the cube root of the quadratic expression $x^2-8x+20$. The goal is to find the value(s) of $x$ that make this equation true. To do this, we'll need to isolate $x$ by reversing the operations applied to it. This involves understanding how to deal with exponents and quadratic expressions. Let's start by tackling the cube root.

Step 1: Eliminating the Cube Root

The first step in solving this equation is to eliminate the cube root. To do this, we need to raise both sides of the equation to the power of 3. This will effectively undo the cube root operation. So, we have:

$(x^2-8x+20)^{\frac{1}{3}}=2$

Raise both sides to the power of 3:

$((x^2-8x+20)^{\frac{1}{3}})^3 = 2^3$

This simplifies to:

$x^2-8x+20 = 8$

Now we have a quadratic equation, which is much easier to work with. The next step is to rearrange this equation into the standard quadratic form.

Step 2: Rearranging the Equation

Now that we've eliminated the cube root, we have a quadratic equation: $x^2-8x+20 = 8$. To solve this, we need to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. This will allow us to use various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. Let's subtract 8 from both sides of the equation:

$x^2 - 8x + 20 - 8 = 8 - 8$

This simplifies to:

$x^2 - 8x + 12 = 0$

Now we have our quadratic equation in the standard form. The next step is to solve this equation for $x$.

Step 3: Solving the Quadratic Equation

We have the quadratic equation $x^2 - 8x + 12 = 0$. There are several ways to solve this quadratic equation. We can try factoring, completing the square, or using the quadratic formula. In this case, factoring seems like the most straightforward approach. We need to find two numbers that multiply to 12 and add up to -8. These numbers are -6 and -2. So, we can factor the quadratic equation as follows:

$(x - 6)(x - 2) = 0$

Now, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for $x$:

$x - 6 = 0$ or $x - 2 = 0$

Solving these equations gives us:

$x = 6$ or $x = 2$

So, we have two potential solutions for $x$. The next step is to check these solutions to make sure they are valid.

Step 4: Checking the Solutions

We found two potential solutions for $x$: $x = 6$ and $x = 2$. To ensure these solutions are valid, we need to check the solutions by plugging them back into the original equation $(x^2-8x+20)^{\frac{1}{3}}=2$. Let's start with $x = 6$:

$(6^2 - 8(6) + 20)^{\frac{1}{3}} = (36 - 48 + 20)^{\frac{1}{3}} = (8)^{\frac{1}{3}} = 2$

So, $x = 6$ is a valid solution. Now let's check $x = 2$:

$(2^2 - 8(2) + 20)^{\frac{1}{3}} = (4 - 16 + 20)^{\frac{1}{3}} = (8)^{\frac{1}{3}} = 2$

So, $x = 2$ is also a valid solution. Both solutions satisfy the original equation.

Conclusion

In this article, we successfully solved the equation $(x^2-8x+20)^{\frac{1}{3}}=2$. We started by understanding the equation and then systematically worked through the steps to isolate $x$. We eliminated the cube root, rearranged the equation into a standard quadratic form, solved the quadratic equation by factoring, and finally, checked our solutions to ensure they were valid. The solutions we found are $x = 6$ and $x = 2$. This problem demonstrates the power of algebraic manipulation and equation-solving techniques. Remember, practice makes perfect, so keep solving equations and challenging yourself!

For further learning and practice on similar mathematical problems, you might find resources at Khan Academy helpful.