Solving Radical Equations: Find The Value Of Y

Hey guys! Let's dive into a fun math problem: solving radical equations! We're gonna tackle an equation that involves square roots, and our mission is to find the value of y. Don't worry, it's not as scary as it sounds! We'll break it down step by step, making it super easy to understand. So, grab your pencils and let's get started. The equation we're working with is: $\sqrt{-9y + 4} = \sqrt{-4y + 1}$. Our goal is crystal clear: find out what y equals. This process involves a few key steps that we'll explore together. We'll start by isolating the radicals, and then we'll square both sides to get rid of those pesky square roots. From there, it's just a matter of simplifying and solving for y. And finally, we'll check our answer to make sure everything's on the up and up. Ready? Let's go!

Understanding Radical Equations

First off, what exactly is a radical equation? Well, it's simply an equation where the variable we're trying to find is under a radical sign – in our case, a square root. These equations can sometimes be a bit tricky because we have to remember the rules of radicals. When we solve these kinds of equations, we might sometimes end up with solutions that don't actually work in the original equation. These are called extraneous solutions. That's why it's super important to always double-check our answers at the end. In our equation $\sqrt{-9y + 4} = \sqrt{-4y + 1}$, the square root is the radical. The expression under the square root, like $-9y + 4$ and $-4y + 1$, is called the radicand. To solve the equation, we need to isolate the radical and remove it. The primary method for doing this is to square both sides. This eliminates the square roots and lets us deal with a regular algebraic equation. So, keep an eye out for those extraneous solutions; they're the sneaky ones that can mess us up if we're not careful. Think of it like this: We're detectives, and we need to verify if the solution is a real one or just a decoy.

Squaring Both Sides and Simplifying

Alright, let's get down to business. Our equation is $\sqrt{-9y + 4} = \sqrt{-4y + 1}$. The first step is to get rid of those square roots. How do we do that? By squaring both sides of the equation! When we square a square root, they cancel each other out. So, squaring both sides gives us:

$(\sqrt{-9y + 4})^2 = (\sqrt{-4y + 1})^2$

This simplifies to:

$-9y + 4 = -4y + 1$

See how much cleaner that looks? Now we have a simple linear equation. Our next step is to get all the y terms on one side and the constants on the other side. Let's add $9y$ to both sides and subtract $1$ from both sides. This gives us:

$4 - 1 = -4y + 9y$

Which simplifies to:

$3 = 5y$

Now, to solve for y, we just need to divide both sides by 5:

$y = \frac{3}{5}$

There you have it! We've found a value for y. But, hold on a sec! We're not completely done yet. We still need to check if this solution actually works in the original equation. Remember those extraneous solutions we talked about? Let's not skip the important step of substituting this value back into the original equation.

Checking the Solution

Okay, guys, here's the most important part: Checking our solution. We found that $y = \frac{3}{5}$. Now, let's substitute this value back into the original equation, $\sqrt{-9y + 4} = \sqrt{-4y + 1}$, to see if it holds true. This is like a final test to make sure our answer is correct. Let's substitute $\frac{3}{5}$ into the equation:

$\sqrt{-9(\frac{3}{5}) + 4} = \sqrt{-4(\frac{3}{5}) + 1}$

Let's simplify each side step by step.

First, for the left side:

$-9(\frac{3}{5}) = -\frac{27}{5}$

So, we have:

$\sqrt{-\frac{27}{5} + 4}$

To add the numbers, we need a common denominator, which is 5. So, we change 4 to $\frac{20}{5}$:

$\sqrt{-\frac{27}{5} + \frac{20}{5}} = \sqrt{-\frac{7}{5}}$

Wait a minute! We have a negative number under the square root. That means this is not a real number. Let's check the right side.

Now, for the right side:

$-4(\frac{3}{5}) = -\frac{12}{5}$

So, we have:

$\sqrt{-\frac{12}{5} + 1}$

We need a common denominator again, so we change 1 to $\frac{5}{5}$:

$\sqrt{-\frac{12}{5} + \frac{5}{5}} = \sqrt{-\frac{7}{5}}$

Again, we have a negative number under the square root. Therefore, the square root on both sides is undefined because we cannot take the square root of a negative number in the realm of real numbers. This indicates that our initial value is not correct.

Why the Solution Doesn't Work & Revisiting the Problem

Alright, folks, so here's the deal. When we substituted $y = \frac{3}{5}$ back into the original equation, we ran into a problem: we ended up with negative numbers under the square root. Remember, the square root of a negative number isn't a real number. It's an imaginary number, and we're not dealing with those here. So, what does this mean? It means our solution, $y = \frac{3}{5}$, doesn't actually work in the original equation. It's an extraneous solution. This can happen when solving radical equations because squaring both sides can introduce extra solutions that don't satisfy the original equation. We have to be very careful to check our answers to make sure they're valid.

So, does that mean there's no solution? Not necessarily! It could mean that there's no real number solution. Let's go back and examine our work. We squared both sides of the equation to eliminate the square roots, which is a valid operation. We solved the resulting linear equation correctly. The mistake, if any, often lies in overlooking the initial conditions or the restrictions that come with square roots. For a square root to be valid, the expression inside the square root (the radicand) must be greater than or equal to zero. So, we must have $-9y + 4 \ge 0$ and $-4y + 1 \ge 0$. Let's solve these inequalities and see what values of y are valid for our original equation. The issue here is the properties of the square root function, which we sometimes overlook. We may also have to check our assumptions and ensure we haven’t made any errors in our calculations. Remember to stay alert and double-check those calculations!

Correcting the Approach: Checking Domain Restrictions

Since we ran into trouble with our initial solution, let's take a more rigorous approach. When dealing with radical equations, it's essential to consider the domain of the equation. This means we need to determine the values of y for which the expressions inside the square roots are non-negative. This will ensure that our final solution is a real number and not an extraneous one. Let's revisit the original equation: $\sqrt{-9y + 4} = \sqrt{-4y + 1}$.

First, for the left side, we need $-9y + 4 \ge 0$. Solving for y:

$-9y \ge -4$

Dividing both sides by -9 (and remembering to flip the inequality sign because we're dividing by a negative number):

$y \le \frac{4}{9}$

Next, for the right side, we need $-4y + 1 \ge 0$. Solving for y:

$-4y \ge -1$

Dividing both sides by -4 (and flipping the inequality sign):

$y \le \frac{1}{4}$

Now, for a solution to be valid, it must satisfy both of these conditions: $y \le \frac{4}{9}$ and $y \le \frac{1}{4}$. Since $\frac{1}{4}$ is less than $\frac{4}{9}$, the more restrictive condition is $y \le \frac{1}{4}$. This tells us that any valid solution must be less than or equal to $\frac{1}{4}$. Let's also go back and check our algebraic manipulations. Sometimes a small arithmetic error can lead us astray.

Finding the Correct Solution (or Lack Thereof)

Let's re-examine our earlier steps to find out if there's a solution that fits the equation. Remember, we initially squared both sides and simplified to get $y = \frac{3}{5}$. But we quickly found that this did not work when checked back into the original equation. Because of the conditions of the square roots, we can see that no real solution exists. Since $\frac{3}{5}$ is not less than or equal to $\frac{1}{4}$, the value we got earlier doesn't fit within the range. The value of y must satisfy both of these conditions. Since the initial solution, y=3/5, does not satisfy the conditions, then there is no solution for the given equation.

Therefore, we can confidently say that there's no real number solution to this radical equation. The reason is that any value of y that we might find through our algebraic manipulations will either make the expressions inside the square roots negative (which isn't allowed in the real number system) or will not satisfy the original equation. It's a bit like trying to find a treasure that doesn't exist – you can search all you want, but you won't find it!

Conclusion

Alright, guys, we've come to the end of our adventure with this radical equation! We've learned that even though we can solve it algebraically, we have to be super careful about checking our answers. Remember to check those domain restrictions by ensuring the value under the square root is non-negative and to always substitute your solution back into the original equation. We've seen how easy it is to find a solution that seems correct, only to realize later that it doesn't work. That's why checking your answer is so crucial.

So, the final answer to this equation is that there is no real solution. Keep practicing, keep checking your work, and you'll become a pro at solving radical equations in no time! Keep up the great work and always remember to double-check those solutions! You've got this!