Solving Equations: A Step-by-Step Guide

Hey guys! Today, we're diving deep into the fascinating world of mathematics, specifically focusing on how to tackle equations. You know, those puzzles where you need to find the unknown value of 'x'. We've got a neat example here that shows us the steps to solve a particular equation, and by the end of this article, you'll be a pro at understanding these kinds of problems. So, grab your thinking caps, and let's get this party started!

The Equation at Hand

Our main quest today is to solve the equation: $\sqrt{30-2x} = x-3$. This might look a little intimidating with that square root hanging around, but don't sweat it! The key is to isolate 'x', and the steps provided break it down beautifully. We're going to dissect each stage, making sure we understand why each step is taken. This isn't just about getting the answer, it's about understanding the process, which is super crucial in math. We'll explore the logic behind squaring both sides, why we get a quadratic equation, and how to find those potential solutions. Stick around, and we'll unravel this mystery together. This is going to be fun, I promise!

Step 1: Squaring Both Sides

Alright, team, let's talk about the first major move in solving our equation $\sqrt{30-2x} = x-3$. The biggest hurdle here is that pesky square root. In mathematics, a common strategy to eliminate a square root is to square both sides of the equation. Think of it like this: if you have a number and you want to undo the square root, you square it. So, we apply this to our equation. Squaring the left side, $\sqrt{30-2x}$, simply gives us $30-2x$. On the right side, we have $(x-3)^2$. Now, this is where some of you might get a little stuck. Remember your algebra! $(x-3)^2$ is not $x^2 - 9$. It's $(x-3)(x-3)$. We need to foil this out, meaning First, Outer, Inner, Last:

• First: $x * x = x^2$
• Outer: $x * -3 = -3x$
• Inner: $-3 * x = -3x$
• Last: $-3 * -3 = 9$

Adding these together, we get $x^2 - 3x - 3x + 9$, which simplifies to $x^2 - 6x + 9$. So, after squaring both sides, our equation transforms from $\sqrt{30-2x} = x-3$ into $30-2x = x^2 - 6x + 9$. This step is fundamental because it gets rid of the square root, allowing us to work with a more familiar polynomial equation. It's like clearing the path so you can move forward more easily. Keep this in mind, guys, because this technique of squaring both sides is a lifesaver for many radical equations.

Step 2: Rearranging into a Quadratic Equation

Now that we've successfully squared both sides and have the equation $30-2x = x^2 - 6x + 9$, our next mission is to get it into a standard form that we can easily work with. Most of the time in math, when we deal with equations involving $x^2$ terms, we aim to rearrange them into the standard quadratic form, which is $ax^2 + bx + c = 0$. To do this, we need to move all the terms to one side of the equation, leaving zero on the other side. It doesn't matter which side you choose, but traditionally, we move terms to the side where the $x^2$ term is positive, which in this case is the right side.

Let's start by moving the $30$ from the left side to the right. We do this by subtracting $30$ from both sides:

$30 - 2x - 30 = x^2 - 6x + 9 - 30$

This simplifies to:

$-2x = x^2 - 6x - 21$

Next, we need to move the $-2x$ term. We add $2x$ to both sides:

$-2x + 2x = x^2 - 6x - 21 + 2x$

This gives us:

$0 = x^2 - 4x - 21$

And there you have it! We've successfully transformed our original radical equation into a standard quadratic equation: $x^2 - 4x - 21 = 0$. This is a huge step because quadratic equations have well-established methods for solving them, such as factoring or using the quadratic formula. Recognizing this form is key to progressing. So, remember, when you see that $x^2$ term, try to get everything on one side to equal zero. It opens up a whole new set of tools for you to use, making the problem much more manageable. Keep this strategy in your back pocket, it's a real game-changer!

Step 3: Factoring the Quadratic Equation

We've reached a pivotal point, guys! We now have the quadratic equation $x^2 - 4x - 21 = 0$. The next logical step is to find the values of $x$ that make this equation true. One of the most efficient ways to solve a quadratic equation is by factoring, especially if the numbers are friendly. Factoring means rewriting the quadratic expression as a product of two linear expressions (binomials). We are looking for two numbers that multiply to give us the constant term ($c = -21$) and add up to give us the coefficient of the $x$ term ($b = -4$).

Let's brainstorm. We need two numbers that multiply to $-21$. Possible pairs are:

• $1$ and $-21$
• $-1$ and $21$
• $3$ and $-7$
• $-3$ and $7$

Now, let's check which of these pairs adds up to $-4$:

• $1 + (-21) = -20$ (Nope)
• $-1 + 21 = 20$ (Nope)
• $3 + (-7) = -4$ (Bingo!)
• $-3 + 7 = 4$ (Close, but not quite)

So, the magic pair of numbers is $3$ and $-7$. This means we can factor our quadratic expression $x^2 - 4x - 21$ into $(x+3)(x-7)$.

Therefore, our equation $x^2 - 4x - 21 = 0$ becomes $(x+3)(x-7) = 0$. This form is incredibly useful because for the product of two things to be zero, at least one of them must be zero. This leads us to our potential solutions. Factoring is a super powerful skill in algebra. It allows you to break down complex expressions into simpler ones, making them easier to solve. Practice this skill, and you'll find many quadratic equations become surprisingly easy to handle. It's all about finding those right numbers that fit the puzzle!

Step 4: Finding the Possible Solutions

We're in the home stretch, folks! We've factored our quadratic equation down to $(x+3)(x-7) = 0$. Remember our rule: if the product of two factors is zero, then at least one of the factors must be zero. This gives us two separate, simpler equations to solve:

1. $x+3 = 0$
2. $x-7 = 0$

Solving the first equation, $x+3 = 0$, we simply subtract $3$ from both sides to get $x = -3$.

Solving the second equation, $x-7 = 0$, we add $7$ to both sides to get $x = 7$.

So, based on our factoring, the possible solutions to the equation are $x = -3$ and $x = 7$. It's crucial to use the word