Solving \(\sqrt{9y - 18} = Y\) A Step-by-Step Guide

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Hey guys! Let's dive into solving an exciting algebraic equation where we need to find the value of ${ y }$ that satisfies ${\sqrt{9y - 18} = y}$. This type of equation, involving square roots, often requires a bit of algebraic maneuvering and a keen eye to ensure we capture all valid solutions while avoiding extraneous ones. Buckle up; it’s going to be a mathematical adventure!

Understanding the Equation

Before we jump into the solving process, let’s first understand what our equation, ${\sqrt{9y - 18} = y}$, is telling us. We have a square root on one side and a simple variable, ${ y }$, on the other. The key thing to remember with square roots is that the expression inside the square root (the radicand) must be non-negative, and the result of a square root is also non-negative. This gives us our first crucial condition: ${ 9y - 18 \ge 0 }$.

Initial Constraints and Domain

Initial constraints are critical. Let's break down why this radicand needs to be non-negative. Think about it: the square root of a negative number isn't a real number, and we're looking for real number solutions for ${ y }$. So, ${ 9y - 18 }$ has to be greater than or equal to zero. This also implies that ${ y }$ itself must be non-negative since it's equal to the square root, which always yields a non-negative value.

To find the exact constraint on ${ y }$, we solve the inequality:

${ 9y - 18 \ge 0 }$

Add 18 to both sides:

${ 9y \ge 18 }$

Divide by 9:

${ y \ge 2 }$

So, this is super important! We know that any solution for ${ y }$ must be greater than or equal to 2. This is our initial condition, and it will help us filter out any extraneous solutions later on. This condition stems directly from the domain of the square root function involved.

Steps to Solve the Equation

Now, let's get into the actual solving. Here's how we can tackle the equation ${\sqrt{9y - 18} = y}$ step by step.

1. Squaring Both Sides

To eliminate the square root, the most straightforward approach is to square both sides of the equation. When we do this, we get:

${ (\sqrt{9y - 18})^2 = y^2 }$

This simplifies to:

${ 9y - 18 = y^2 }$

Squaring both sides transforms the equation into a more manageable form—a quadratic equation. Remember, though, that squaring can sometimes introduce solutions that don't actually satisfy the original equation (extraneous solutions), which is why we have that initial condition to help us!

2. Rearranging into a Quadratic Equation

Okay, so now we have ${ 9y - 18 = y^2 }$. To solve this, we want to rearrange it into the standard form of a quadratic equation, which is ${ ay^2 + by + c = 0 }$. Let's subtract ${ 9y }$ and add 18 to both sides to get everything on one side:

${ 0 = y^2 - 9y + 18 }$

Or, rewriting it:

${ y^2 - 9y + 18 = 0 }$

Now we have a quadratic equation ready to be factored or solved using the quadratic formula.

3. Factoring the Quadratic Equation

Factoring is often the quickest way to solve a quadratic equation if the factors are easily discernible. We need to find two numbers that multiply to 18 and add up to -9. Think about it: what two numbers fit that description? How about -3 and -6? Yep, ${ (-3) \times (-6) = 18 }$ and ${ -3 + (-6) = -9 }$. So, we can factor the quadratic equation as follows:

${ (y - 3)(y - 6) = 0 }$

4. Finding Potential Solutions

Now that we have factored the quadratic equation, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. So, we set each factor equal to zero and solve for ${ y }$:

${ y - 3 = 0 \quad \text{or} \quad y - 6 = 0 }$

Solving these gives us two potential solutions:

${ y = 3 \quad \text{or} \quad y = 6 }$

5. Checking for Extraneous Solutions

Alright, we've got two potential solutions, but we're not out of the woods yet! Remember that initial condition we found, ${ y \ge 2 }$? And remember how squaring both sides can sometimes introduce extraneous solutions? This is where that initial constraint comes into play. We need to check if both ${ y = 3 }$ and ${ y = 6 }$ actually satisfy the original equation ${\sqrt{9y - 18} = y}$.

Checking ${ y = 3 }$

Plug ${ y = 3 }$ into the original equation:

${ \sqrt{9(3) - 18} = 3 }$

Simplify:

${ \sqrt{27 - 18} = 3 }$

${ \sqrt{9} = 3 }$

${ 3 = 3 }$

So, ${ y = 3 }$ checks out! It's a valid solution.

Checking ${ y = 6 }$

Now let's plug in ${ y = 6 }$:

${ \sqrt{9(6) - 18} = 6 }$

Simplify:

${ \sqrt{54 - 18} = 6 }$

${ \sqrt{36} = 6 }$

${ 6 = 6 }$

Guess what? ${ y = 6 }$ also works! It's another valid solution.

Final Solutions

After all that algebraic footwork and careful checking, we've found that both ${ y = 3 }$ and ${ y = 6 }$ are indeed solutions to the equation ${\sqrt{9y - 18} = y}$. We made sure to consider the domain restrictions and checked for extraneous solutions, so we can confidently say that these are the real deal.

Wrapping Up

Solving equations involving square roots can be a bit like navigating a maze, but with a systematic approach and attention to detail, we can find our way to the correct solutions. Remember, the key steps are:

1. Understanding and setting up initial constraints.
2. Squaring both sides to eliminate the square root.
3. Rearranging the equation into a standard form (like a quadratic).
4. Solving for potential solutions.
5. Crucially checking for extraneous solutions.

By following these steps, you'll be well-equipped to tackle similar problems and emerge victorious! Keep practicing, and you'll become a master equation solver in no time. Awesome job, guys! We nailed it!