Solving Equations: Substitution Method Explained

Hey guys! Today, we're diving into the fascinating world of equation-solving, specifically focusing on a technique called substitution. This method can be a real game-changer when you're faced with equations that look a bit intimidating at first glance. We'll break down the process step-by-step, making it super easy to understand and apply. Let's tackle the equation $2 x^{\frac{2}{3}}-15 x^{\frac{1}{3}}-8=0$ together and see how substitution can work its magic.

Understanding the Equation and Why Substitution?

Before we jump into the solution, let's take a closer look at the equation: $2 x^{\frac{2}{3}}-15 x^{\frac{1}{3}}-8=0$. Notice those fractional exponents? They might seem a little scary, but don't worry! The key here is to recognize a pattern. We have terms with $x^{\frac{2}{3}}$ and $x^{\frac{1}{3}}$. This is a classic setup for using substitution. Substitution helps us simplify complex equations by replacing a part of the equation with a single variable, making it easier to manipulate and solve. Think of it as a clever way to transform a tricky problem into a more familiar one. By using substitution, we can transform this equation into a quadratic equation, which we know how to solve.

Why is this important? Well, quadratic equations have a well-established set of methods for finding solutions (like factoring, the quadratic formula, etc.). By converting our original equation into a quadratic form, we can leverage these techniques to find the values of x that satisfy the equation. So, the goal here is not just to solve this particular equation, but also to learn a powerful technique that can be applied to a variety of similar problems. Recognizing when to use substitution is a crucial skill in algebra and calculus.

Breaking Down the Fractional Exponents

Let's quickly recap what fractional exponents mean. The exponent $\frac{1}{3}$ means the cube root, so $x^{\frac{1}{3}}$ is the same as $\sqrt[3]{x}$. Similarly, $x^{\frac{2}{3}}$ can be written as $(x^{\frac{1}{3}})^2$ or $(\sqrt[3]{x})^2$. This understanding is crucial for making the right substitution. We can see that $x^{\frac{2}{3}}$ is simply the square of $x^{\frac{1}{3}}$. This relationship is what makes substitution such an effective technique in this case. By recognizing this pattern, we can introduce a new variable that represents $x^{\frac{1}{3}}$, and the equation will become much more manageable. Remember, the beauty of math lies in spotting these connections and using them to simplify problems.

The Substitution Step: Making the Equation Simpler

Alright, let's get our hands dirty and make the substitution. This is where the magic happens! We're going to introduce a new variable, let's call it u, to represent $x^{\frac{1}{3}}$. So, we have: $u = x^{\frac{1}{3}}$. Now, what happens to $x^{\frac{2}{3}}$? Well, as we discussed earlier, $x^{\frac{2}{3}}$ is the same as $(x^{\frac{1}{3}})^2$. Since $u = x^{\frac{1}{3}}$, we can say that $x^{\frac{2}{3}} = u^2$. See how things are starting to simplify?

Now, we can rewrite our original equation, $2 x^{\frac{2}{3}}-15 x^{\frac{1}{3}}-8=0$, in terms of u. Replacing $x^{\frac{2}{3}}$ with $u^2$ and $x^{\frac{1}{3}}$ with u, we get: $2u^2 - 15u - 8 = 0$. Ta-da! Look at that! We've transformed our equation with those fractional exponents into a good old quadratic equation. This is a major step forward because we have a toolbox full of methods for solving quadratic equations.

Why This Works: The Power of Transformation

Substitution is a powerful technique because it allows us to transform a complex problem into a simpler one that we already know how to solve. In this case, we turned a non-quadratic equation into a quadratic equation. This transformation doesn't change the solutions; it just makes them easier to find. It's like taking a tangled ball of yarn and carefully unraveling it so you can see the individual strands. By making the right substitution, we can untangle the equation and reveal its underlying structure.

Solving the Quadratic Equation

Now that we have the quadratic equation $2u^2 - 15u - 8 = 0$, we can solve it using several methods. Factoring is often the quickest way if the quadratic factors nicely. The quadratic formula is a more general method that always works, even when factoring is difficult. And completing the square is another useful technique that can be particularly helpful in certain situations.

Let's try factoring first. We need to find two numbers that multiply to $(2)(-8) = -16$ and add up to -15. Those numbers are -16 and 1. So, we can rewrite the middle term, -15u, as -16u + u: $2u^2 - 16u + u - 8 = 0$. Now, we can factor by grouping: $2u(u - 8) + 1(u - 8) = 0$. This gives us $(2u + 1)(u - 8) = 0$. Setting each factor equal to zero, we get $2u + 1 = 0$ or $u - 8 = 0$. Solving for u, we find $u = -\frac{1}{2}$ or $u = 8$.

Factoring vs. Quadratic Formula: Choosing the Right Tool

We successfully factored this quadratic equation, but what if it hadn't been so easy? That's where the quadratic formula comes in handy. The quadratic formula is a foolproof method for solving any quadratic equation of the form $ax^2 + bx + c = 0$. The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. If you're ever unsure about factoring, or if the quadratic looks particularly messy, the quadratic formula is your best friend.

Back-Substitution: Finding the Values of x

We've found the values of u, but remember, we're trying to solve for x. This is where the back-substitution comes in. We need to go back to our original substitution, $u = x^{\frac{1}{3}}$, and replace u with the values we just found.

So, we have two cases:

1. When $u = -\frac{1}{2}$: We have $-\frac{1}{2} = x^{\frac{1}{3}}$. To solve for x, we need to cube both sides of the equation: $(-\frac{1}{2})^3 = (x^{\frac{1}{3}})^3$. This gives us $x = -\frac{1}{8}$.
2. When $u = 8$: We have $8 = x^{\frac{1}{3}}$. Again, we cube both sides: $8^3 = (x^{\frac{1}{3}})^3$. This gives us $x = 512$.

Therefore, the solutions to the original equation are $x = -\frac{1}{8}$ and $x = 512$.

The Importance of Back-Substitution

Back-substitution is a crucial step in solving equations using substitution. It's easy to get caught up in solving for the new variable and forget to go back to the original variable. Always remember to back-substitute to find the solutions in terms of the original variable. This ensures that you've actually answered the question that was asked.

Verifying the Solutions

It's always a good idea to check your solutions by plugging them back into the original equation. This helps you catch any errors you might have made along the way. Let's plug $x = -\frac{1}{8}$ and $x = 512$ back into the equation $2 x^{\frac{2}{3}}-15 x^{\frac{1}{3}}-8=0$ to see if they work.

For $x = -\frac{1}{8}$:

$2(-\frac{1}{8})^{\frac{2}{3}} - 15(-\frac{1}{8})^{\frac{1}{3}} - 8 = 2(\frac{1}{4}) - 15(-\frac{1}{2}) - 8 = \frac{1}{2} + \frac{15}{2} - 8 = \frac{16}{2} - 8 = 8 - 8 = 0$. So, $x = -\frac{1}{8}$ is a valid solution.

For $x = 512$:

$2(512)^{\frac{2}{3}} - 15(512)^{\frac{1}{3}} - 8 = 2(64) - 15(8) - 8 = 128 - 120 - 8 = 0$. So, $x = 512$ is also a valid solution.

Why Verification Matters

Verifying your solutions is like double-checking your work on a test. It's a simple step that can save you from making mistakes. By plugging your solutions back into the original equation, you can ensure that they actually satisfy the equation. This is especially important when dealing with equations that involve radicals or fractional exponents, as these can sometimes lead to extraneous solutions (solutions that don't actually work).

Conclusion: Mastering Substitution

So, there you have it! We've successfully solved the equation $2 x^{\frac{2}{3}}-15 x^{\frac{1}{3}}-8=0$ using the method of substitution. We saw how substitution can transform a complex equation into a simpler one, making it easier to solve. Remember the key steps:

1. Identify the pattern: Look for repeating expressions or relationships between terms in the equation.
2. Make the substitution: Introduce a new variable to represent the repeating expression.
3. Solve the new equation: Solve the simplified equation for the new variable.
4. Back-substitute: Replace the new variable with its original expression and solve for the original variable.
5. Verify the solutions: Plug the solutions back into the original equation to check your work.

By mastering the technique of substitution, you'll be well-equipped to tackle a wide range of equations. Keep practicing, and you'll become a pro at solving even the most challenging problems! This method isn't just useful for these types of equations, but it's a fundamental technique that pops up in higher-level math like calculus, so getting comfortable with it now will pay off big time later. Keep up the great work, guys!