Solve $8 X^{rac{2}{3}}-63 X^{rac{1}{3}}-8=0$: A Math Guide

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Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of algebraic equations, specifically tackling a problem that might look a little intimidating at first glance: $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$. Don't let those fractional exponents scare you away! This equation, while appearing complex, can be elegantly solved using a technique that transforms it into a more familiar quadratic form. We'll break down each step, ensuring you understand the logic behind the solution, making it accessible even if you're not a seasoned mathematician. Our goal here is to demystify this type of equation, providing you with the tools and confidence to approach similar problems in the future. We'll explore the substitution method, which is key to unlocking the solution, and walk through the process of finding the values of $x$ that satisfy this equation. So, grab your favorite thinking cap, and let's embark on this mathematical journey together. We promise it will be an enlightening experience, shedding light on how to manipulate and solve equations that involve fractional powers. By the end of this article, you'll not only have the answer but also a solid understanding of the principles involved, ready to tackle more challenging problems with ease.

Understanding Fractional Exponents and the Equation Structure

Before we jump into solving, let's take a moment to appreciate the components of our equation: $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$. The key feature here is the presence of fractional exponents, specifically $x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$. It's crucial to recognize the relationship between these terms. Notice that $x^{\frac{2}{3}}$ can be rewritten as $(x^{\frac{1}{3}})^2$. This observation is the lynchpin that allows us to simplify the equation. When you see an exponent that is double another exponent within the same equation, a common strategy is to use a substitution. This is where the magic happens, transforming a seemingly complex expression into a standard quadratic equation, which we are typically more comfortable with. Think of it like this: if you have a shape you don't recognize, you might try to see if it fits into a mold you already know. That's precisely what we're doing here. We're taking this equation, which has a quadratic-like structure due to the exponents, and fitting it into the familiar $ay^2 + by + c = 0$ form. This method is incredibly powerful because it applies to a wide range of equations with similar exponent relationships, not just this specific one. By understanding this fundamental property of exponents, you unlock a powerful problem-solving technique that can be used across various mathematical contexts. We'll delve into the substitution itself in the next section, but it's this relationship between $x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$ that truly paves the way for a straightforward solution.

The Power of Substitution: Transforming the Equation

Now, let's harness the power of substitution to simplify $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$. As we identified, $x^{\frac{2}{3}}$ is equivalent to $(x^{\frac{1}{3}})^2$. This allows us to introduce a new variable. Let's set $y = x^{\frac{1}{3}}$. If $y = x^{\frac{1}{3}}$, then $y^2 = (x^{\frac{1}{3}})^2 = x^{\frac{2}{3}}$. Substituting these into our original equation, we get: $8y^2 - 63y - 8 = 0$. See? Just like that, we've transformed our equation with fractional exponents into a standard quadratic equation! This is a common and highly effective technique in algebra. Whenever you encounter an equation where one term's variable exponent is exactly double another term's variable exponent, consider using a substitution like this. It simplifies the problem considerably, allowing you to use familiar methods for solving quadratic equations. The beauty of this substitution is that it bridges the gap between more complex exponential forms and the well-understood quadratic form. It's like finding a secret passage that leads you from an unfamiliar territory to a place where you feel completely at home. The new equation, $8y^2 - 63y - 8 = 0$, is now in the standard form $ay^2 + by + c = 0$, where $a=8$, $b=-63$, and $c=-8$. This form is something most algebra students are very familiar with, and we have several reliable methods to solve it. We'll explore these methods in the next section, paving the way to finding the values of $y$, which will then allow us to find the values of $x$. This step is critical, as it represents the successful simplification of the initial complex problem into a manageable one.

Solving the Quadratic Equation for 'y'

With our transformed equation $8y^2 - 63y - 8 = 0$, we can now employ standard methods for solving quadratic equations. There are a few popular approaches: factoring, completing the square, or using the quadratic formula. For this particular equation, factoring is a viable and often quickest method if the numbers align. Let's try to factor it. We are looking for two numbers that multiply to $a imes c$ (which is $8 imes -8 = -64$) and add up to $b$ (which is $-63$). The numbers that fit this description are $-64$ and $1$. So, we can rewrite the middle term: $8y^2 - 64y + 1y - 8 = 0$. Now, we can factor by grouping: $8y(y - 8) + 1(y - 8) = 0$. Notice that we have a common factor of $(y - 8)$. Factoring this out, we get: $(8y + 1)(y - 8) = 0$. For this product to be zero, at least one of the factors must be zero. Therefore, we have two possible solutions for $y$:

1. 8y + 1 = 0 ightarrow 8y = -1 ightarrow y = -rac{1}{8}
2. $y - 8 = 0 ightarrow y = 8$

So, our solutions for $y$ are $y = 8$ and y = -rac{1}{8}. If factoring felt a bit tricky, the quadratic formula is always a reliable fallback. The quadratic formula for an equation $ay^2 + by + c = 0$ is y = rac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Plugging in our values ($a=8$, $b=-63$, $c=-8$):

y = rac{-(-63) \pm \sqrt{(-63)^2 - 4(8)(-8)}}{2(8)}

y = rac{63 \pm \sqrt{3969 + 256}}{16}

y = rac{63 \pm \sqrt{4225}}{16}

y = rac{63 \pm 65}{16}

This gives us two solutions:

$y_1 = \frac{63 + 65}{16} = \frac{128}{16} = 8$

$y_2 = \frac{63 - 65}{16} = \frac{-2}{16} = -\frac{1}{8}$

Both methods yield the same results for $y$: $8$ and -rac{1}{8}. These are the intermediate solutions that will help us find the final values for $x$. The ability to successfully solve this quadratic step is crucial, as it directly leads us to the potential solutions for the original, more complex equation.

Finding the 'x' Values: Reversing the Substitution

We've successfully found the possible values for $y$, which are $y = 8$ and y = -rac{1}{8}. Now, we need to reverse our substitution to find the values of $x$. Remember, we initially set $y = x^{\frac{1}{3}}$. So, we'll take each of our $y$ values and set it equal to $x^{\frac{1}{3}}$:

Case 1: $y = 8$

$x^{\frac{1}{3}} = 8$

To solve for $x$, we need to cube both sides of the equation:

$(x^{\frac{1}{3}})^3 = 8^3$

$x = 512$

Case 2: y = -rac{1}{8}

x^{\frac{1}{3}} = -rac{1}{8}

Again, we cube both sides to isolate $x$:

(x^{\frac{1}{3}})^3 = ig(-rac{1}{8}ig)^3

x = -rac{1}{512}

So, the solutions for $x$ are $512$ and -rac{1}{512}. It's always a good practice to check these solutions by plugging them back into the original equation, $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$, to ensure they hold true. This verification step confirms the accuracy of our work and reinforces our understanding of how the substitution method leads us to the correct answers. The process of reversing the substitution is as important as the substitution itself, as it brings us back to the original variable and provides the ultimate answer to the problem.

Verification: Ensuring the Solutions are Correct

To be absolutely sure our solutions are accurate for the equation $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$, let's verify them by plugging them back into the original equation. This step is critical in mathematics to catch any errors and confirm that our derived values truly satisfy the given conditions.

Checking $x = 512$:

First, let's find $x^{\frac{1}{3}}$ and $x^{\frac{2}{3}}$ for $x=512$.

$x^{\frac{1}{3}} = 512^{\frac{1}{3}} = \sqrt[3]{512} = 8$

$x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2 = 8^2 = 64$

Now, substitute these values into the original equation:

$8(64) - 63(8) - 8 = 0$

$512 - 504 - 8 = 0$

$8 - 8 = 0$

$0 = 0$

This is true, so $x=512$ is a valid solution.

Checking x = -rac{1}{512}:

Now, let's find $x^{\frac{1}{3}}$ and $x^{\frac{2}{3}}$ for x=-rac{1}{512}.

x^{\frac{1}{3}} = ig(-rac{1}{512}ig)^{\frac{1}{3}} = \sqrt[3]{-\frac{1}{512}} = -rac{1}{8}

x^{\frac{2}{3}} = (x^{\frac{1}{3}})^2 = ig(-rac{1}{8}ig)^2 = rac{1}{64}

Substitute these values into the original equation:

8ig(rac{1}{64}ig) - 63ig(-rac{1}{8}ig) - 8 = 0

rac{8}{64} + rac{63}{8} - 8 = 0

rac{1}{8} + rac{63}{8} - rac{64}{8} = 0

rac{1 + 63 - 64}{8} = 0

rac{64 - 64}{8} = 0

rac{0}{8} = 0

$0 = 0$

This is also true, confirming that x=-rac{1}{512} is another valid solution.

Both solutions have been verified, demonstrating the effectiveness of the substitution method and the accuracy of our calculations. This thorough verification process is an essential part of problem-solving in mathematics, ensuring that the answers we arrive at are correct and reliable. It instills confidence in our abilities and provides a clear path forward when encountering similar algebraic challenges.

Conclusion: Mastering Equations with Fractional Exponents

We've successfully navigated the intricacies of the equation $8 x^{\frac{2}{3}}-63 x^{\frac{1}{3}}-8=0$, arriving at the solutions $x=512$ and x=-rac{1}{512}. The key takeaway from this problem is the power of strategic substitution. By recognizing that $x^{\frac{2}{3}}$ is the square of $x^{\frac{1}{3}}$ and introducing a new variable, $y = x^{\frac{1}{3}}$, we transformed a complex equation into a manageable quadratic form. This technique is invaluable for solving a wide array of equations that feature relationships between variable exponents. Remember, when you encounter an equation where one exponent is double another, think substitution! This approach simplifies the problem, allowing you to leverage familiar algebraic methods like factoring or the quadratic formula. The journey from fractional exponents to a standard quadratic and back again highlights the interconnectedness of mathematical concepts. Practice is, of course, essential. The more you work through problems like this, the more intuitive these techniques will become. Don't shy away from equations that initially seem daunting; often, a clever algebraic manipulation can reveal a straightforward path to the solution. We hope this detailed walkthrough has demystified equations with fractional exponents and equipped you with a valuable problem-solving strategy. Keep exploring, keep practicing, and you'll find yourself confidently tackling even more challenging mathematical puzzles. For further exploration into algebraic techniques and equation solving, you might find resources on Khan Academy or MathWorld incredibly helpful.