Mastering Radical Equations: Solve $3\sqrt{x+3}-\sqrt{x-2}=7$

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Welcome to the World of Radical Equations!

Hey guys, ever stared down a math problem that looked like a tangled mess of square roots and numbers, making you think, "Ugh, where do I even begin with this?" Well, if you've been wrestling with equations like our main event today – the formidable $3\sqrt{x+3}-\sqrt{x-2}=7$ – then you've come to the absolute right place! Radical equations, which are essentially equations where the variable hides beneath a square root (or sometimes a cube root, etc.), can seem super intimidating at first. They've got a reputation for being a bit tricky, mostly because of those sneaky square roots and the potential for what we call extraneous solutions. But don't you worry, because by the time you finish this article, you'll not only understand how to solve this specific beast of an equation, but you'll also gain some seriously powerful problem-solving skills that apply to any radical equation you might encounter.

Our mission here isn't just about finding the 'x' in this particular puzzle; it's about equipping you with a clear, step-by-step strategy. We're going to dive deep into understanding domain restrictions (super important, trust me!), meticulously walk through each algebraic manipulation, and then, most crucially, learn the art of checking our solutions to make sure we don't fall for any mathematical trickery. I'll break down everything in a friendly, conversational tone, just like we're chilling and solving this together. This isn't just about memorizing steps; it's about truly understanding the "why" behind each move, so you can build real, lasting algebra skills and confidently conquer any radical challenge thrown your way. So, buckle up, grab a pen and paper, and let's turn you into a radical equation master!

Getting Down to Basics: What Exactly Are Radical Equations?

Before we jump headfirst into solving $3\sqrt{x+3}-\sqrt{x-2}=7$, let's quickly chat about what radical equations actually are and why they need a special kind of love and attention. Simply put, a radical equation is any equation that contains a variable under a radical symbol, usually a square root. For example, $\sqrt{x}=5$ is a radical equation. Easy, right? But the moment things get a little more complex, like with multiple radical terms or other expressions, that's when we need a solid game plan. The biggest, most fundamental rule when dealing with square roots (and indeed, any even-indexed root) is the concept of the domain. You see, you cannot take the square root of a negative number in the realm of real numbers. So, for any expression like $\sqrt{A}$, the term 'A' must be greater than or equal to zero (i.e., $A \ge 0$). This isn't just a suggestion; it's an unbreakable law of mathematics!

For our specific equation, $3\sqrt{x+3}-\sqrt{x-2}=7$, this domain restriction immediately tells us two things: First, the expression inside the first square root, $x+3$, must be non-negative. So, $x+3 \ge 0$, which means $x \ge -3$. Second, the expression inside the second square root, $x-2$, must also be non-negative. So, $x-2 \ge 0$, which means $x \ge 2$. For both conditions to be true simultaneously, our 'x' must be greater than or equal to 2. So, right off the bat, we know any solution we find must satisfy $x \ge 2$. If it doesn't, it's an instant imposter, an extraneous solution, and we can toss it out without a second thought! This first step, checking the domain, is super important and often overlooked, but it's your first line of defense against incorrect answers.

Now, how do we get rid of those pesky square roots? By squaring both sides of the equation! This is our main tool. However, squaring both sides comes with a significant caveat: it can introduce those aforementioned extraneous solutions. Think about it: if you have $x=3$, then $x^2=9$. If you have $x=-3$, then $x^2=9$ too. When we work backward from $x^2=9$ to $x=\pm3$, we see that squaring can hide the original sign. So, when we square both sides of an equation with radicals, we might accidentally introduce solutions that don't actually satisfy the original equation. This is precisely why checking your solutions is not optional but an absolutely crucial, non-negotiable final step in solving radical equations. Trust me, skipping the check is like forgetting your parachute when skydiving – don't do it!

Your Step-by-Step Guide to Crushing $3\sqrt{x+3}-\sqrt{x-2}=7$

Alright, guys, this is where the real magic happens! We're going to break down $3\sqrt{x+3}-\sqrt{x-2}=7$ into manageable, bite-sized pieces. Just follow along, and you'll see that even the most complex-looking radical equations yield to a systematic approach. Remember, patience and precision are your best friends here. Let's get started!

Step 1: Isolate a Radical Term

The very first thing you want to do when faced with multiple radicals is to isolate one of them on one side of the equation. This makes the first squaring step much cleaner and helps avoid more complex expansions. Our equation is $3\sqrt{x+3}-\sqrt{x-2}=7$. To isolate one of them, it's usually easiest to move the term with the minus sign. So, let's move $-\sqrt{x-2}$ to the right side of the equation:

$3\sqrt{x+3} = 7 + \sqrt{x-2}$

See? Now we have one radical all by itself on the left side. This is crucial because when we square it, the radical will disappear entirely. If we had tried to square the original equation as it was, we would have had to deal with a much more complicated binomial expansion on the left, which would still leave a radical term, and we'd be back to square one (pun intended!).

Step 2: Square Both Sides (Carefully!)

This is a critical juncture where many people make mistakes. Remember what we said about squaring? You have to square the entire side, not just individual terms. So, we'll square both the left and right sides of our isolated equation:

$(3\sqrt{x+3})^2 = (7 + \sqrt{x-2})^2$

Let's tackle the left side first. $(3\sqrt{x+3})^2 = 3^2 \cdot (\sqrt{x+3})^2 = 9(x+3) = 9x+27$.

Now for the right side. This is a binomial, so we must use the $(a+b)^2 = a^2 + 2ab + b^2$ formula. Here, $a=7$ and $b=\sqrt{x-2}$:

$(7 + \sqrt{x-2})^2 = 7^2 + 2(7)(\sqrt{x-2}) + (\sqrt{x-2})^2 = 49 + 14\sqrt{x-2} + (x-2) = x + 47 + 14\sqrt{x-2}$.

Putting it all together, our equation now looks like this:

$9x+27 = x+47+14\sqrt{x-2}$

Notice that we still have one radical term left. That's perfectly normal for equations with two initial radicals! We'll just repeat the isolation process.

Step 3: Isolate the Remaining Radical

Our goal now is to get that $14\sqrt{x-2}$ term by itself. To do this, we'll move all the non-radical terms from the right side to the left side:

$9x+27 - x - 47 = 14\sqrt{x-2}$

Combine like terms on the left:

$8x - 20 = 14\sqrt{x-2}$

Now, before we square again, let's look at that equation: $8x-20 = 14\sqrt{x-2}$. See how all the coefficients ($8$, $20$, and $14$) are even numbers? We can simplify this by dividing the entire equation by 2. This step is a small but mighty tip; it makes the numbers smaller and reduces the chances of errors in the next squaring step:

$4x - 10 = 7\sqrt{x-2}$

Much cleaner, right?

Step 4: Square Both Sides Again!

Time for round two of squaring! We've successfully isolated the last radical term, so let's get rid of it. Again, be super careful to square the entire expression on both sides:

$(4x-10)^2 = (7\sqrt{x-2})^2$

Left side: This is another binomial $(a-b)^2 = a^2 - 2ab + b^2$. Here, $a=4x$ and $b=10$:

$(4x-10)^2 = (4x)^2 - 2(4x)(10) + 10^2 = 16x^2 - 80x + 100$.

Right side: $(7\sqrt{x-2})^2 = 7^2 \cdot (\sqrt{x-2})^2 = 49(x-2) = 49x - 98$.

So, our equation has transformed into a quadratic equation:

$16x^2 - 80x + 100 = 49x - 98$

Woohoo! No more radicals! We're on the home stretch to finding our potential solutions.

Step 5: Solve the Quadratic Equation

Now we have a standard quadratic equation. To solve it, we need to set it equal to zero. Let's move all the terms from the right side to the left side:

$16x^2 - 80x - 49x + 100 + 98 = 0$

Combine the like terms:

$16x^2 - 129x + 198 = 0$

This is a quadratic equation of the form $ax^2 + bx + c = 0$. While you could try factoring, the numbers are a bit large, so the quadratic formula is usually a reliable friend here: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a=16$, $b=-129$, and $c=198$.

Let's first calculate the discriminant, $b^2 - 4ac$:

$(-129)^2 - 4(16)(198) = 16641 - 12672 = 3969$.

Now, let's find the square root of the discriminant: $\sqrt{3969} = 63$. (Phew! A nice whole number, which means our roots are rational.)

Plug these values into the quadratic formula:

$x = \frac{-(-129) \pm 63}{2(16)} = \frac{129 \pm 63}{32}$

This gives us two potential solutions:

$x_1 = \frac{129 + 63}{32} = \frac{192}{32} = 6$

$x_2 = \frac{129 - 63}{32} = \frac{66}{32} = \frac{33}{16}$

So, we have two candidates for our solution: $x=6$ and $x=\frac{33}{16}$. But remember, we're not done yet! We must check these solutions in the original equation to see if they're valid or if one is an extraneous solution.

The Moment of Truth: Checking Our Roots

Alright, guys, you've done the heavy lifting, the algebraic gymnastics are complete, and you've got two shiny potential solutions: $x=6$ and $x=\frac{33}{16}$. But remember our cardinal rule for radical equations? We absolutely, positively, 100% must check these roots in the original equation ($3\sqrt{x+3}-\sqrt{x-2}=7$). This is where we separate the true solutions from the imposter extraneous solutions that sometimes sneak in when we square both sides of an equation. Don't skip this step, ever! This is where all your hard work either pays off or helps you correctly identify a false lead. Let's do it!

First, let's recall our domain restriction from the beginning: $x \ge 2$. Both of our potential solutions, $x=6$ and $x=\frac{33}{16}$ (which is approximately $2.0625$), satisfy this condition. So, they both have the potential to be valid. Now, for the full check:

Checking $x=6$:

Let's substitute $x=6$ back into the original equation: $3\sqrt{x+3}-\sqrt{x-2}=7$.

$3\sqrt{6+3} - \sqrt{6-2}$

$3\sqrt{9} - \sqrt{4}$

We know that $\sqrt{9}=3$ and $\sqrt{4}=2$. So, this becomes:

$3(3) - 2$

$9 - 2$

$7$

Does $7=7$? YES, IT DOES! This is fantastic! This means that $x=6$ is a valid solution to our equation. Give yourself a pat on the back for that one!

Checking $x=\frac{33}{16}$:

Now, let's plug $x=\frac{33}{16}$ into the original equation: $3\sqrt{x+3}-\sqrt{x-2}=7$.

$3\sqrt{\frac{33}{16}+3} - \sqrt{\frac{33}{16}-2}$

To make this easier, let's convert the whole numbers to fractions with a denominator of 16:

$3 = \frac{3 \cdot 16}{16} = \frac{48}{16}$

$2 = \frac{2 \cdot 16}{16} = \frac{32}{16}$

Now substitute these back:

$3\sqrt{\frac{33}{16}+\frac{48}{16}} - \sqrt{\frac{33}{16}-\frac{32}{16}}$

Simplify the terms inside the square roots:

$3\sqrt{\frac{33+48}{16}} - \sqrt{\frac{33-32}{16}}$

$3\sqrt{\frac{81}{16}} - \sqrt{\frac{1}{16}}$

Now, take the square roots:

$3\left(\frac{\sqrt{81}}{\sqrt{16}}\right) - \left(\frac{\sqrt{1}}{\sqrt{16}}\right)$

$3\left(\frac{9}{4}\right) - \frac{1}{4}$

Multiply the first term:

$\frac{27}{4} - \frac{1}{4}$

Subtract the fractions:

$\frac{26}{4}$

Simplify the fraction:

$\frac{13}{2}$

Now, the big question: Does $\frac{13}{2} = 7$? Well, $\frac{13}{2}$ is $6.5$. And $6.5 \ne 7$. NO, IT DOES NOT!

This means that $x=\frac{33}{16}$ is an extraneous solution. It appeared during our algebraic steps because of the squaring process, but it doesn't satisfy the original equation. Therefore, we must discard it.

The Final Answer:

After all that hard work and meticulous checking, we can confidently say that the only true solution to the equation $3\sqrt{x+3}-\sqrt{x-2}=7$ is $x=6$.

Common Pitfalls and Pro Tips for Radical Equations

Alright, you've conquered a tough one, which means you're already ahead of the game! But solving radical equations can sometimes feel like navigating a minefield, with potential traps lurking at every turn. I've been there, guys, and I've made (and seen!) all the common mistakes. So, let me drop some serious pro tips to help you avoid those pitfalls and cruise through future problems with confidence. These aren't just suggestions; they're habits that will save you time, frustration, and points on your next exam. Let's dive into some wisdom that goes beyond just the steps.

First and foremost, never, ever forget the domain! Seriously, I can't stress this enough. Before you even move a single term, quickly establish the domain restrictions. For square roots, the expression under the radical must be non-negative. For $3\sqrt{x+3}-\sqrt{x-2}=7$, we found $x \ge 2$. By knowing this from the start, you can immediately rule out any potential solutions that fall outside this range, saving you time during the checking phase. It's your first layer of defense against extraneous solutions.

My second crucial tip: when squaring a binomial, use the formula correctly! This is perhaps the most common error. Remember, $(a+b)^2 = a^2 + 2ab + b^2$, and not just $a^2+b^2$. Similarly, $(a-b)^2 = a^2 - 2ab + b^2$. When we squared $(7+\sqrt{x-2})^2$, it wasn't $49 + (x-2)$; it was $49 + 14\sqrt{x-2} + (x-2)$. That middle term, $2ab$, is a total game-changer and omitting it will lead you down a completely wrong path. Always take your time expanding these, maybe even write out the FOIL method if it helps: $(a+b)(a+b)$.

Third, always try to isolate a radical before squaring. This is why our first step was $3\sqrt{x+3} = 7 + \sqrt{x-2}$. If you try to square an equation with multiple radicals on one side, like $(3\sqrt{x+3}-\sqrt{x-2})^2$, you'll get a massive, complicated mess with cross-terms that still contain radicals! It's an unnecessary headache. One radical at a time, guys. That's the mantra. It makes the algebra significantly cleaner and reduces the likelihood of errors.

Fourth, and this bears repeating because of its sheer importance: CHECK. ALL. SOLUTIONS. Yes, all of them! I know, it adds an extra step at the end, and sometimes you might be tempted to skip it, especially if you're feeling confident. But as we saw with $x=\frac{33}{16}$, even perfectly calculated algebraic solutions can turn out to be extraneous. Plug each potential root back into the original equation to confirm its validity. This is your ultimate safeguard against incorrect answers in radical equations.

Finally, simplify your equations whenever possible. Remember when we had $8x-20 = 14\sqrt{x-2}$? We divided everything by 2 to get $4x-10 = 7\sqrt{x-2}$. This simple step makes the numbers smaller, the squaring easier, and the quadratic equation much more manageable. Look for common factors to divide out, or combine like terms promptly. These little simplifications add up to a smoother problem-solving experience. Like any skill, solving equations with square roots gets easier with practice. Don't be discouraged by initial difficulties; every mistake is a learning opportunity. Keep practicing, and you'll become a true pro!

Wrapping It Up: Your Newfound Radical Equation Superpowers!

Wow, you guys totally rocked it! We started with an equation that might have seemed like a daunting puzzle, $3\sqrt{x+3}-\sqrt{x-2}=7$, and together, we systematically broke it down, solved it, and most importantly, understood why each step was necessary. You've not just solved one specific problem; you've gained a comprehensive toolkit for tackling any radical equation that comes your way. That's a serious upgrade to your algebra skills, and something to be really proud of!

Let's do a quick recap of the superpowers you've unlocked today. Remember the essential workflow for mastering these equations:

1. Establish the Domain: Always start by identifying the values of 'x' for which the radicals are defined (i.e., the expressions under the even roots must be non-negative). This helps you filter out invalid solutions early on.
2. Isolate a Radical: Get one radical term all by itself on one side of the equation. This simplifies the squaring process immensely.
3. Square Both Sides (Carefully!): This is your main tool for eliminating radicals. Be absolutely meticulous when squaring binomials – don't forget that crucial middle term from $(a+b)^2$.
4. Repeat as Necessary: If you still have radicals after the first squaring, just repeat steps 2 and 3 until all radicals are gone.
5. Solve the Resulting Equation: Once the radicals are gone, you'll usually be left with a linear or quadratic equation. Apply your existing algebraic prowess to find the potential solutions.
6. Crucially, Check All Potential Solutions! This is the non-negotiable final step. Plug every single candidate solution back into the original equation. This is how you identify and discard those pesky extraneous solutions that may have been introduced by squaring. Only the values that satisfy the original equation are true solutions.

By following these steps, you've not only found that $x=6$ is the unique solution to our challenging equation, but you've also internalized the discipline and critical thinking required for solving equations with square roots. This isn't just about math problems; it's about building a structured approach to problem-solving that's valuable in all areas of life. So, go forth and conquer, my friends! Practice makes perfect, so keep those math muscles flexing, and you'll be a true master of radical equations in no time. You got this!"