Solving The Equation $-7+\left(x^2-19\right)^{\frac{3}{4}}=20$ A Step-by-Step Guide

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Hey guys! Let's dive into solving this equation together. We've got a bit of a tricky one here, but don't worry, we'll break it down step by step. Our mission is to find the solutions to the equation: $-7+\left(x2-19\right){\frac{3}{4}}=20$

In this article, we will explore the methods used to solve such equations, focusing on understanding each step in detail. Math can seem intimidating at first, but with a clear, step-by-step approach, even complex problems become manageable. We will begin by isolating the term with the exponent, and then carefully undoing the fractional exponent. Along the way, we'll discuss why certain operations are valid and what potential pitfalls to watch out for, like extraneous solutions. Our goal is not just to find the answers, but to understand why they are the correct answers.

Step 1: Isolating the Exponential Term

The first thing we want to do is isolate the term that has the exponent. This means we need to get (x2−19)34\left(x^2-19\right)^{\frac{3}{4}} all by itself on one side of the equation. To do this, we'll add 7 to both sides of the equation. This is a crucial step in simplifying the equation and making it easier to work with. When we add the same number to both sides of an equation, we maintain the equality, which is a fundamental principle in algebra. So, let’s get started by adding 7 to both sides:

−7+(x2−19)34+7=20+7-7 + \left(x^2 - 19\right)^{\frac{3}{4}} + 7 = 20 + 7

This simplifies to:

(x2−19)34=27\left(x^2 - 19\right)^{\frac{3}{4}} = 27

Now we have the exponential term isolated, which sets us up perfectly for the next step. Isolating terms is a common strategy in solving equations, allowing us to focus on the part that needs further manipulation. By isolating the exponential term, we've made the equation look much cleaner and more approachable. It's like clearing away the clutter before starting a project – it helps us see what we need to do next more clearly. We can now see the structure of the equation more easily, making the subsequent steps flow logically. This careful preparation is often the key to successfully solving mathematical problems.

Step 2: Eliminating the Fractional Exponent

Now comes the fun part – getting rid of that fractional exponent! We have (x2−19)34=27\left(x^2 - 19\right)^{\frac{3}{4}} = 27. To eliminate the exponent 34\frac{3}{4}, we need to raise both sides of the equation to the power of its reciprocal, which is 43\frac{4}{3}. This is a neat trick, because when you raise a power to its reciprocal, the exponents multiply to 1, effectively canceling each other out. Remember, whatever we do to one side of the equation, we have to do to the other side to keep things balanced. This principle is at the heart of solving equations in algebra.

So, let's raise both sides to the power of 43\frac{4}{3}:

[(x2−19)34]43=2743\left[\left(x^2 - 19\right)^{\frac{3}{4}}\right]^{\frac{4}{3}} = 27^{\frac{4}{3}}

On the left side, the exponents 34\frac{3}{4} and 43\frac{4}{3} multiply to 1, leaving us with:

x2−19=2743x^2 - 19 = 27^{\frac{4}{3}}

On the right side, we need to evaluate 274327^{\frac{4}{3}}. Remember, a fractional exponent like 43\frac{4}{3} means we're taking both a root and a power. The denominator (3) tells us to take the cube root, and the numerator (4) tells us to raise the result to the fourth power. So, we first find the cube root of 27, which is 3 (since 33=273^3 = 27). Then, we raise 3 to the fourth power: 34=813^4 = 81. Therefore, 2743=8127^{\frac{4}{3}} = 81.

Now our equation looks much simpler:

x2−19=81x^2 - 19 = 81

By carefully applying the reciprocal exponent, we've transformed a complex-looking equation into a much more manageable form. This step showcases the power of understanding exponent rules in simplifying algebraic expressions. It also highlights the importance of breaking down a problem into smaller, easier-to-handle parts. We're now one step closer to finding the solutions for x.

Step 3: Solving for x²

Alright, we're making great progress! We've simplified the equation to x2−19=81x^2 - 19 = 81. Our next step is to isolate x2x^2. To do this, we'll add 19 to both sides of the equation. Just like before, adding the same value to both sides keeps the equation balanced and maintains the equality. This is a fundamental principle in algebra and is crucial for solving equations correctly. By isolating x2x^2, we're getting closer to finding the values of x that satisfy the original equation. This step is a classic example of how algebraic manipulations can help us unravel a problem and move towards a solution.

Let's add 19 to both sides:

x2−19+19=81+19x^2 - 19 + 19 = 81 + 19

This simplifies to:

x2=100x^2 = 100

Now we have a much simpler equation to work with: x2=100x^2 = 100. We're just one step away from finding the solutions for x. Isolating x2x^2 is a key step because it allows us to directly address the square and find the values that, when squared, give us 100. This process of simplification is what makes algebra so powerful – we can take a complex equation and, through careful steps, reduce it to a form that's easy to solve.

Step 4: Finding the Values of x

We're in the home stretch now! We've got x2=100x^2 = 100. To find the values of x, we need to take the square root of both sides of the equation. Now, this is a crucial point: when we take the square root, we need to remember that there are both positive and negative solutions. Why? Because both a positive number and its negative counterpart, when squared, will give us the same positive result. For example, both 10210^2 and (−10)2(-10)^2 equal 100. This is a common place where people sometimes miss a solution, so let's make sure we get it right!

Taking the square root of both sides gives us:

x=±100x = \pm\sqrt{100}

So, we have two possible solutions:

x=10orx=−10x = 10 \quad \text{or} \quad x = -10

It's super important to consider both positive and negative roots when solving equations like this. By remembering this detail, we ensure that we capture all possible solutions. We're not just looking for a solution, but all solutions that satisfy the equation. This careful attention to detail is what makes our approach thorough and reliable. We've successfully found two potential solutions for x, but our work isn't quite done yet. We need to do one more crucial step to make sure these solutions are valid.

Step 5: Checking for Extraneous Solutions

Okay, we've found two potential solutions: x=10x = 10 and x=−10x = -10. But before we celebrate, we need to do a super important check: we need to make sure these solutions actually work in the original equation. Why? Because when we raise both sides of an equation to a power, especially a fractional power, we can sometimes introduce what are called extraneous solutions. These are values that we get as solutions, but they don't actually satisfy the original equation. It's like finding a key that fits the lock, but doesn't actually open the door.

So, let's take each potential solution and plug it back into the original equation: $-7+\left(x2-19\right){\frac{3}{4}}=20$

First, let's check x=10x = 10:

−7+(102−19)34=−7+(100−19)34=−7+(81)34-7 + \left(10^2 - 19\right)^{\frac{3}{4}} = -7 + \left(100 - 19\right)^{\frac{3}{4}} = -7 + \left(81\right)^{\frac{3}{4}}

Now, 813481^{\frac{3}{4}} means we take the fourth root of 81 (which is 3) and then cube it: 33=273^3 = 27. So we have:

−7+27=20-7 + 27 = 20

This checks out! So, x=10x = 10 is a valid solution.

Now, let's check x=−10x = -10:

−7+((−10)2−19)34=−7+(100−19)34=−7+(81)34-7 + \left((-10)^2 - 19\right)^{\frac{3}{4}} = -7 + \left(100 - 19\right)^{\frac{3}{4}} = -7 + \left(81\right)^{\frac{3}{4}}

Notice anything? This is exactly the same as when we plugged in x=10x = 10. We already know that 8134=2781^{\frac{3}{4}} = 27, so we have:

−7+27=20-7 + 27 = 20

This also checks out! So, x=−10x = -10 is also a valid solution.

By taking the time to check our solutions, we've ensured that we're not including any extraneous solutions. This step is a hallmark of careful and thorough problem-solving. It’s like double-checking your work before you submit it – you want to be sure you've got the right answer. In this case, both of our potential solutions turned out to be valid, which is fantastic!

Conclusion: The Solutions We Found

Alright guys, we did it! We successfully navigated through the equation $-7+\left(x2-19\right){\frac{3}{4}}=20$ and found the solutions. By carefully isolating the exponential term, eliminating the fractional exponent, solving for x2x^2, finding the square roots, and – most importantly – checking for extraneous solutions, we arrived at our final answer.

We found that the solutions to the equation are x=10x = 10 and x=−10x = -10. These are the values that, when plugged back into the original equation, make the equation true. It's a testament to the power of methodical problem-solving and the importance of understanding each step along the way. Math isn't just about finding the right answer; it's about understanding why that answer is correct.

So, to recap, the correct answers are:

  • -10
  • 10

We tackled a tricky equation and came out on top. Great job, everyone! Remember, the key to mastering math is practice, patience, and a willingness to break down problems into smaller, more manageable steps. Keep up the great work, and you'll be solving even more complex equations in no time!