Solving Equations With Radicals A Step By Step Guide

Hey guys! Let's dive into solving equations with radicals, specifically focusing on how to find the right values to make an equation true. We're going to break down a problem where we need to drag values into the correct spots in an equation. Not all values will be used, which adds a little extra challenge. So, grab your thinking caps, and let’s get started!

Understanding the Problem

Our main goal here is to find the values that make a given equation true. This often involves a mix of algebraic manipulation and careful selection of numbers. In this particular case, we have an equation with a cube root and some given values to choose from. The twist? Not all values will fit, meaning we need to be strategic about our choices.

The equation we're tackling looks something like this:

$$\sqrt[3]{6y} - 14 = ?$$

And we have a set of values:

$$\begin{array}{llllll}101 & 8 & 17 & 28 & 3 & 4\end{array}$$

We need to pick two values from this set and place them correctly in the equation to make it true, keeping in mind that $y \neq 0$.

Why This Matters

Understanding how to solve equations like this is super important for a few reasons. First off, it's a fundamental concept in algebra. Mastering these skills helps you tackle more complex problems later on. Plus, it's not just about math class; these problem-solving skills come in handy in all sorts of real-life situations. Whether you're calculating finances, planning a project, or even just figuring out a puzzle, the ability to think logically and manipulate numbers is a huge asset.

Breaking Down the Basics

Before we jump into the specifics of this problem, let's quickly review some key concepts. Remember, when we're solving equations, we're essentially trying to isolate the variable. This means getting the variable (in our case, y) all by itself on one side of the equation. We do this by performing the same operations on both sides of the equation to maintain balance. For example, if we subtract a number from one side, we need to subtract the same number from the other side.

Radicals, like the cube root in our equation, are just another way of expressing exponents. The cube root of a number is the value that, when multiplied by itself three times, equals the original number. So, the cube root of 8 is 2 because 2 * 2 * 2 = 8. Knowing this helps us simplify equations containing radicals.

Now that we've got the basics covered, let's roll up our sleeves and dive into solving the problem!

Step-by-Step Solution

Okay, guys, let's get our hands dirty and solve this equation step by step. Remember, our equation looks like this:

$$\sqrt[3]{6y} - 14 = ?$$

And we have the following values to play with:

$$\begin{array}{llllll}101 & 8 & 17 & 28 & 3 & 4\end{array}$$

Our mission is to find two values from this set that fit into the equation and make it true. Let's break it down.

Step 1: Isolate the Cube Root

First things first, we want to get that cube root term by itself on one side of the equation. To do this, we need to get rid of the -14. How do we do that? By adding 14 to both sides of the equation. This is a classic move in equation solving – doing the same thing to both sides to keep everything balanced.

So, our equation becomes:

$$\sqrt[3]{6y} - 14 + 14 = ? + 14$$

Simplifying, we get:

$$\sqrt[3]{6y} = ? + 14$$

Step 2: Eliminate the Cube Root

Now we've got the cube root isolated, but we need to deal with it to get to y. The opposite of taking the cube root is cubing, so we're going to cube both sides of the equation. This means raising each side to the power of 3.

$$(\sqrt[3]{6y})^3 = (? + 14)^3$$

This simplifies to:

$$6y = (? + 14)^3$$

Step 3: Substitute and Solve

Here's where the fun begins! We need to find a value from our set that, when added to 14, results in a number that, when cubed, gives us a multiple of 6. This might sound tricky, but let's take it one step at a time.

Let's try the value 17 from our set. If we put 17 in place of the question mark, we get:

$$6y = (17 + 14)^3$$

$$6y = (31)^3$$

$$6y = 29791$$

Now, we divide both sides by 6 to solve for y:

$$y = \frac{29791}{6}$$

This gives us a non-integer value for y, which isn't ideal since we're looking for simpler solutions. So, 17 might not be the right fit.

Let's try another value. How about 101?

$$6y = (101 + 14)^3$$

$$6y = (115)^3$$

$$6y = 1520875$$

Dividing by 6:

$$y = \frac{1520875}{6}$$

Again, we get a non-integer value for y. It seems we need a value that, when added to 14 and then cubed, results in a number that's nicely divisible by 6.

Let's try 4:

$$6y = (4 + 14)^3$$

$$6y = (18)^3$$

$$6y = 5832$$

Now, divide by 6:

$$y = \frac{5832}{6}$$

$$y = 972$$

Aha! We got a whole number for y. This is a good sign. So, one of our values is 4.

Step 4: Find the Corresponding Value for y

Now that we know one value is 4, let's plug it back into our simplified equation:

$$\sqrt[3]{6y} = 4 + 14$$

$$\sqrt[3]{6y} = 18$$

Cube both sides:

$$6y = 18^3$$

$$6y = 5832$$

Divide by 6:

$$y = 972$$

So, when the right side of the equation equals 18 (which is 4 + 14), y equals 972. Now, let's plug y back into the original equation to find the value that goes on the left side.

Step 5: Solve for the Remaining Value

Our original equation was:

$$\sqrt[3]{6y} - 14 = ?$$

We know y = 972, so let's substitute that in:

$$\sqrt[3]{6 \times 972} - 14 = ?$$

$$\sqrt[3]{5832} - 14 = ?$$

Since we know that $18^3 = 5832$, the cube root of 5832 is 18:

$$18 - 14 = ?$$

$$4 = ?$$

Wait a second! This means that the value on the right side of the original equation should be 4, but we already used 4 as one of our values to add to 14. This seems like a bit of a loop, but it actually confirms that we're on the right track.

Let’s double-check our work.

We used 4 on the right side after isolating the cube root:

$$\sqrt[3]{6y} = 4 + 14$$

$$\sqrt[3]{6y} = 18$$

Then we cubed both sides:

$$6y = 18^3$$

$$6y = 5832$$

$$y = 972$$

Now, plugging y = 972 back into the original equation:

$$\sqrt[3]{6 \times 972} - 14 = ?$$

$$\sqrt[3]{5832} - 14 = ?$$

$$18 - 14 = ?$$

$$4 = ?$$

So, the equation becomes:

$$\sqrt[3]{6 \times 972} - 14 = 4$$

This confirms that our values are correct. We used 4 and 18 (which is 4 + 14). But remember, we need two values from our original set. We used 4, and we found that adding 14 to it gave us 18, which is the cube root we needed. So, the second value we need is the result of our original equation: 4.

But hold on! We need to make sure we are picking from our original set of numbers: $\begin{array}{llllll}101 & 8 & 17 & 28 & 3 & 4\end{array}$. We already identified that 4 is one of the values. The next value we need to identify is what the cube root expression should equal after adding 14 to the unknown value. Let’s revisit this step:

$$\sqrt[3]{6y} = ? + 14$$

We determined that when we plug in 4 for the question mark, we get:

$$\sqrt[3]{6y} = 4 + 14$$

$$\sqrt[3]{6y} = 18$$

So, $ \sqrt[3]{6y}$ must equal 18. Now we go back to the original equation:

$$\sqrt[3]{6y} - 14 = ?$$

We substitute 18 for $ \sqrt[3]{6y}$:

$$18 - 14 = ?$$

$$4 = ?$$

So, the two values that make the equation true are 4 (the value of the expression) and 4 (the value we add to 14 to get the cube root).

Step 6: State the Solution

So, after all that brain-busting work, we've found our values! The two values that make the equation true are 4 and 18.

Common Pitfalls and How to Avoid Them

Alright, guys, let's talk about some common mistakes that people make when solving equations like this, so you can steer clear of them. It's always good to learn from others' slip-ups, right?

Mistake #1: Forgetting to Do the Same Thing to Both Sides

This is a classic blunder. Remember, when you're solving equations, you're essentially trying to keep things balanced. If you add, subtract, multiply, or divide on one side, you must do the exact same thing on the other side. Otherwise, you're messing with the equation's equilibrium, and your answer will be off.

How to Avoid It: Always double-check that you've performed the same operation on both sides of the equation. It might even help to write it out explicitly, like “Add 14 to both sides” or “Divide both sides by 6.”

Mistake #2: Messing Up the Order of Operations

Ah, PEMDAS (or BODMAS, depending on where you went to school). This little acronym is your best friend when solving equations. It reminds you to tackle parentheses/brackets first, then exponents/orders, then multiplication and division (from left to right), and finally addition and subtraction (also from left to right). Ignoring this order can lead to major errors.

How to Avoid It: Write out each step clearly, and always refer back to PEMDAS to make sure you're doing things in the correct order. It might feel a bit tedious at first, but it'll save you from making mistakes in the long run.

Mistake #3: Incorrectly Dealing with Radicals

Radicals can be a bit tricky if you're not careful. One common mistake is forgetting that the opposite of a cube root is cubing (raising to the power of 3), and so on. Another is messing up the simplification process.

How to Avoid It: Take your time when dealing with radicals. Make sure you understand what the radical means (e.g., cube root means “what number multiplied by itself three times gives you this?”). And when you're trying to eliminate a radical, remember to do the opposite operation (cubing for cube root, squaring for square root, etc.) on both sides of the equation.

Mistake #4: Not Checking Your Answer

This is a big one! Even if you're feeling super confident about your solution, it's always a good idea to plug it back into the original equation to make sure it works. This is especially important in problems like ours, where not all values will fit.

How to Avoid It: Once you've found a solution, take the extra minute to substitute it back into the original equation. If both sides of the equation are equal, you've got it right! If not, it's time to go back and see where you might have made a mistake.

Mistake #5: Rushing Through the Problem

Math problems, especially ones with multiple steps, require a bit of patience. Rushing through can lead to careless errors, like dropping a negative sign or miscopying a number.

How to Avoid It: Slow down, guys! Take your time, and focus on each step. Write neatly, and double-check your work as you go. It's better to get it right than to finish quickly but make mistakes.

By keeping these pitfalls in mind and practicing good problem-solving habits, you'll be well on your way to conquering even the trickiest equations!

Practice Problems

Okay, guys, now that we've walked through the solution and talked about common mistakes, it's time to put your skills to the test! Practice is key when it comes to mastering equation-solving. So, let's dive into some practice problems that are similar to the one we just tackled.

Practice Problem 1:

Drag the values to the correct location in the equation. Not all values will be used. Which two values will make the equation true, for $y \neq 0$?

$$\begin{array}{llllll}2 & 5 & 11 & 21 & 27 & 64\end{array}$$

$$\sqrt[3]{2y} + 5 = ?$$

Practice Problem 2:

Drag the values to the correct location in the equation. Not all values will be used. Which two values will make the equation true, for $y \neq 0$?

$$\begin{array}{llllll}1 & 7 & 9 & 19 & 36 & 100\end{array}$$

$$\sqrt{y} - 3 = ?$$

Practice Problem 3:

Drag the values to the correct location in the equation. Not all values will be used. Which two values will make the equation true, for $y \neq 0$?

$$\begin{array}{llllll}4 & 6 & 12 & 20 & 24 & 144\end{array}$$

$$\sqrt{3y} - 8 = ?$$

How to Approach These Problems:

1. Isolate the radical: Get the radical term by itself on one side of the equation.
2. Eliminate the radical: Use the opposite operation (squaring for square root, cubing for cube root) on both sides.
3. Substitute and solve: Try different values from the given set until you find one that works.
4. Check your answer: Plug your solution back into the original equation to make sure it's correct.

Remember, practice makes perfect! The more you work through problems like these, the more confident you'll become in your equation-solving abilities.

Conclusion

Alright, guys, we've reached the end of our deep dive into solving equations with radicals! We've covered a lot of ground, from understanding the basic concepts to working through a tricky problem step by step. We've also talked about common pitfalls and how to avoid them, and we've even tackled some practice problems to solidify your skills.

Key Takeaways

Before we wrap up, let's quickly recap the key takeaways from our adventure:

• Isolate the Radical: The first step in solving equations with radicals is to get the radical term by itself on one side of the equation.
• Eliminate the Radical: Use the opposite operation (squaring, cubing, etc.) on both sides to get rid of the radical.
• Solve for the Variable: Once the radical is gone, use algebraic techniques to solve for the variable.
• Check Your Answer: Always, always, always plug your solution back into the original equation to make sure it works.
• Avoid Common Pitfalls: Be mindful of mistakes like forgetting to do the same thing to both sides, messing up the order of operations, and incorrectly dealing with radicals.

Final Thoughts

Solving equations with radicals might seem challenging at first, but with practice and a solid understanding of the steps involved, you'll be able to tackle them with confidence. Remember, math is like any other skill – the more you practice, the better you'll get. So, keep working at it, and don't be afraid to ask for help when you need it.

And remember, guys, math isn't just about getting the right answer; it's about developing problem-solving skills that will serve you well in all areas of life. So, keep challenging yourself, keep learning, and keep having fun with math!

If you have any questions or want to dive deeper into this topic, feel free to leave a comment below. Happy solving!