Unlock The Mystery: Solving $\sqrt{x+66}=x+10$

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Hey there, math enthusiasts and problem-solvers! Ever stared at an equation with a square root and wondered, "How on earth do I get rid of that pesky radical symbol?" Well, you're in the right place, because today, we're going to dive headfirst into solving a super common type of problem: radical equations. Specifically, we're going to tackle the equation $\sqrt{x+66}=x+10$. Don't let the square root intimidate you, guys! By the end of this article, you'll not only know how to solve this exact problem but also understand the why behind each crucial step, empowering you to conquer any radical equation that comes your way. It's all about breaking it down, step by step, and making sure we don't fall for any mathematical tricks along the way. Get ready to flex those brain muscles, because we're about to make some serious math magic happen!

Understanding Radical Equations: What Are We Up Against?

Alright, folks, before we jump into the nitty-gritty of solving $\sqrt{x+66}=x+10$, let's take a moment to really understand what we're dealing with: radical equations. What exactly makes an equation 'radical'? Simply put, it's an equation where the variable (in our case, 'x') is tucked away inside a radical symbol, like a square root ($\sqrt{\phantom{x}}$), a cube root, or any nth root. These equations pop up everywhere in math and science, from physics problems involving distances and velocities to engineering calculations and even in geometry when dealing with the Pythagorean theorem. So, learning how to handle them isn't just about passing a test; it's about building a fundamental skill set that's genuinely useful.

Now, here's the deal: radical equations come with a unique twist that makes them a bit more tricky than your average linear or quadratic equation. The biggest challenge, and what often trips people up, is the potential for extraneous solutions. What are extraneous solutions, you ask? Imagine you're on a treasure hunt, and you follow all the clues perfectly. You dig in two spots, and one spot has the treasure, but the other just has an old boot. The old boot is like an extraneous solution – it's a result you get through correct mathematical operations, but it doesn't actually satisfy the original equation. This often happens when you square both sides of an equation, which is a common and necessary step for eliminating radicals. Squaring both sides can sometimes introduce solutions that weren't valid in the initial setup, because $(-a)^2 = a^2$ means we lose information about the sign. Since the principal square root symbol ($\sqrt{\phantom{x}}$) always denotes a non-negative value, we must be super careful. This means that checking your solutions by plugging them back into the original equation is not just a good habit; it's an absolute must for radical equations. Without this crucial final step, you might end up with an incorrect answer or, worse, multiple incorrect answers. So, keep this golden rule in mind: always verify your solutions! It's the ultimate safeguard against those sneaky extraneous solutions, making sure that your hard work actually leads to the correct solution set for the original problem. We'll explore this in detail as we work through our example, but understanding this fundamental concept upfront will save you a lot of headaches later on. Let's get ready to tackle our specific problem with this crucial knowledge firmly in mind.

The Game Plan: A Step-by-Step Guide to Tackling $\sqrt{x+66}=x+10$

Alright, team, it's time to lay out our strategy for solving $\sqrt{x+66}=x+10$. Think of this as our battle plan, ensuring we approach the problem logically and effectively. Solving radical equations generally follows a specific series of steps designed to systematically eliminate the radical and find the value(s) of 'x'. The core idea is to isolate the radical term, then eliminate it by raising both sides of the equation to the power corresponding to the index of the radical (in our case, squaring for a square root). After that, we'll usually be left with a more familiar type of equation, like a linear or quadratic one, which we can solve using standard algebraic techniques. But remember that super important point we just talked about? The verification step is the non-negotiable final piece of this puzzle. It's what separates a complete, correct solution from one that's potentially flawed with extraneous results.

So, here's our general game plan, which we'll meticulously apply to $\sqrt{x+66}=x+10$:

1. Isolate the Radical Term: This is often the first and most critical step. We need to get the radical expression all by itself on one side of the equation. If there are other terms added or subtracted to the radical, or if it's being multiplied by a constant, our job is to move them to the other side using inverse operations. This prepares the equation for the next step, ensuring that when we raise both sides to a power, we're only targeting the radical itself and not unnecessarily complicating the other side. A clean isolation makes the subsequent algebra much smoother and reduces the chance of errors.
2. Raise Both Sides to the Appropriate Power: Once the radical is isolated, we need to eliminate it. For a square root, we'll square both sides of the equation. If it were a cube root, we'd cube both sides, and so on. This operation effectively 'undoes' the radical. However, this is the step where we must be cautious, as it's the primary source of extraneous solutions. When you square both sides, you're essentially saying that if A=B, then A²=B². While true, it also means that if A=-B, then A²=B² still holds. The original square root notation ($\sqrt{\phantom{x}}$) implies a non-negative result, so we need to ensure our solution respects that original condition.
3. Solve the Resulting Equation: After eliminating the radical, you'll be left with a new equation, typically a linear equation or a quadratic equation (like $ax^2+bx+c=0$). At this stage, you'll use your standard algebraic skills to solve for 'x'. This might involve combining like terms, factoring, using the quadratic formula, or other familiar techniques. The goal here is to find all potential values for 'x' that satisfy this new, non-radical equation.
4. Check All Potential Solutions in the Original Equation: And now, the moment of truth! This is the most crucial step. You must take every potential value of 'x' you found in step 3 and plug it back into the original equation, $\sqrt{x+66}=x+10$. If plugging a value in makes the equation true (i.e., the left side equals the right side), then it's a valid solution. If it makes the equation false, then it's an extraneous solution and must be discarded. Seriously, guys, do not skip this step! It's your ultimate safeguard against getting tripped up by those pesky extraneous results. Without this, your solution might be incomplete or incorrect, and nobody wants that after all this hard work!

With this clear roadmap in hand, we're now perfectly equipped to tackle our specific problem. Let's get started on $\sqrt{x+66}=x+10$ and apply each of these powerful steps to find the correct answer!

Step 1: Isolate the Radical (Wait, It Already Is!)

Okay, so our first step in solving any radical equation is usually to isolate the radical term. This means getting the square root expression all by itself on one side of the equation. For our specific equation, $\sqrt{x+66}=x+10$, we actually have a bit of a head start! Notice that the radical term, $\sqrt{x+66}$, is already isolated on the left side of the equation. There are no other numbers or variables added, subtracted, or multiplied directly to it on that side. This is fantastic because it means we can jump straight to the next step, saving us a little bit of preliminary algebra.

However, it's really important to understand why this step is usually necessary. Imagine if our equation was something like $\sqrt{x+66} - 5 = x+5$. In that scenario, the radical isn't isolated. You'd need to add 5 to both sides first, transforming it into $\sqrt{x+66} = x+10$. Or, if it were $2\sqrt{x+66} = 2x+20$, you'd divide by 2 first. The goal is always to have $\text{radical expression} = \text{everything else}$. Why do we do this? Because when we square both sides (which is our next move), we want to make sure we're only squaring the radical on one side. If we had something like $(\sqrt{x+66} - 5)^2$, we'd have to deal with multiplying out a binomial, which is much more complex and error-prone: $(\sqrt{x+66} - 5)(\sqrt{x+66} - 5) = (x+66) - 10\sqrt{x+66} + 25$. Yikes! That just re-introduces the radical and makes things messier. By isolating the radical first, we ensure that when we square, the radical effectively disappears, simplifying the equation significantly. So, while our current problem is nice and easy because the radical is already isolated, always remember this crucial first move for other, more complex radical equations you might encounter. It's all about setting ourselves up for success!

Step 2: Square Both Sides (Let's Get Rid of That Pesky Root!)

Alright, with our radical term, $\sqrt{x+66}$, beautifully isolated on one side, it's time for the main event: squaring both sides of the equation. This is the most direct way to eliminate the square root symbol and move towards a more familiar algebraic equation. Remember, whatever you do to one side of an equation, you must do to the other side to maintain equality. So, let's take our equation, $\sqrt{x+66}=x+10$, and square both the left and right sides:

$(\sqrt{x+66})^2 = (x+10)^2$

On the left side, squaring a square root is like pressing an 'undo' button; they cancel each other out, leaving us with just the expression inside the radical:

$x+66$

Now, for the right side, $(x+10)^2$, we need to be super careful. This is not simply $x^2+10^2$. Oh no, guys, that's a common mistake! Remember your algebraic identities or just think of it as multiplying a binomial by itself: $(x+10)^2 = (x+10)(x+10)$. To expand this, we use the FOIL method (First, Outer, Inner, Last):

• First: $x \cdot x = x^2$
• Outer: $x \cdot 10 = 10x$
• Inner: $10 \cdot x = 10x$
• Last: $10 \cdot 10 = 100$

Adding these terms together gives us: $x^2 + 10x + 10x + 100 = x^2 + 20x + 100$.

So, after squaring both sides, our equation now transforms into a much friendlier quadratic equation:

$x+66 = x^2 + 20x + 100$

This is a huge step forward! We've successfully eliminated the radical, and now we have a quadratic equation, which we know how to solve. But hold on, remember our earlier discussion about extraneous solutions? This squaring step is precisely where those sneaky imposters can get introduced. When we squared, we effectively turned a potential relationship where $A=|B|$ (since $\sqrt{\text{something}}$ is always non-negative) into $A^2=B^2$. This transformation opens the door for solutions where the original $A=B$ was true, but also for solutions where $A=-B$ would also make $A^2=B^2$ true. That's why the final verification step will be absolutely critical. For now, let's focus on solving this newly formed quadratic equation!

Step 3: Solve the Quadratic Equation (Finding Our Potential Candidates!)

Now that we've successfully squared both sides, we're left with the quadratic equation: $x+66 = x^2 + 20x + 100$. Our goal here is to solve for 'x'. To do this with a quadratic equation, we typically want to set one side of the equation to zero. Let's move all the terms to the right side to keep the $x^2$ term positive, which often makes factoring or using the quadratic formula a bit cleaner. Subtract 'x' from both sides and subtract '66' from both sides:

$0 = x^2 + 20x - x + 100 - 66$

Combine the like terms:

$0 = x^2 + 19x + 34$

Awesome! We now have a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a=1$, $b=19$, and $c=34$. There are a few ways to solve quadratic equations: factoring, using the quadratic formula, or completing the square. For this particular equation, factoring looks promising because we need two numbers that multiply to 34 and add up to 19. Let's list the factors of 34:

• $1 \cdot 34 = 34$
• $2 \cdot 17 = 34$

And behold! $2+17=19$. So, we can factor our quadratic equation as:

$(x+2)(x+17) = 0$

To find the potential solutions for 'x', we set each factor equal to zero:

1. $x+2 = 0 \implies x = -2$
2. $x+17 = 0 \implies x = -17$

So, we have two potential solutions: $x = -2$ and $x = -17$. Fantastic progress! We've done all the hard algebraic work. However, and I cannot stress this enough, these are just potential solutions because of that squaring step we took earlier. We absolutely, positively must verify these in the original equation to see if they are true solutions or if one (or both!) is an extraneous solution. This is the moment of truth, so let's get ready for the ultimate test!

The Ultimate Test: Verifying Our Solutions (Don't Skip This!)

Alright, folks, we've arrived at the most critical stage of solving radical equations: verifying our potential solutions. Remember how we talked about extraneous solutions? This is where we root them out! When you square both sides of an equation, you're essentially losing the information about the sign. For example, if $A^2 = B^2$, it means either $A=B$ or $A=-B$. However, the principal square root symbol, $\sqrt{\phantom{x}}$, by definition, always yields a non-negative result. This means that for our original equation, $\sqrt{x+66}=x+10$, the left side, $\sqrt{x+66}$, must be greater than or equal to zero. Consequently, the right side, $x+10$, must also be greater than or equal to zero for any valid solution. If $x+10$ turns out to be negative for a potential solution, then that solution cannot possibly work because a square root cannot equal a negative number in the realm of real numbers.

Failing to check your answers is the number one reason people get radical equation problems wrong. It's not enough to just solve the quadratic; you have to go back to the source! This isn't just a recommendation; it's a fundamental requirement when dealing with radicals. Let's take our two potential solutions, $x=-2$ and $x=-17$, and put them to the test against the original equation: $\sqrt{x+66}=x+10$.

Checking Solution 1: Is It a Real Deal?

Let's start by plugging $x=-2$ into the original equation: $\sqrt{x+66}=x+10$.

Substitute $x=-2$:

$\sqrt{(-2)+66} = (-2)+10$

Simplify both sides:

$\sqrt{64} = 8$

Now, evaluate the square root:

$8 = 8$

Bingo! The left side equals the right side. This means that $x=-2$ is a valid solution to the original equation. It passed the ultimate test with flying colors! Also, notice that for $x=-2$, the right side, $x+10 = -2+10 = 8$, which is non-negative, satisfying the implicit domain restriction of the principal square root. Always a good sign!

Checking Solution 2: A Wolf in Sheep's Clothing?

Now, let's plug our second potential solution, $x=-17$, into the original equation: $\sqrt{x+66}=x+10$.

Substitute $x=-17$:

$\sqrt{(-17)+66} = (-17)+10$

Simplify both sides:

$\sqrt{49} = -7$

Now, evaluate the square root:

$7 = -7$

Uh oh! This statement is clearly false. $7$ does not equal $-7$. This means that $x=-17$ is an extraneous solution. Even though it came from correct algebraic steps after squaring, it does not satisfy the original equation. Remember how the principal square root symbol $\sqrt{\phantom{x}}$ always denotes the non-negative square root? Here, $\sqrt{49}$ is $7$, not $-7$. For $x=-17$, the right side, $x+10 = -17+10 = -7$, which is a negative value. A principal square root can never equal a negative number in the set of real numbers. Therefore, $x=-17$ is not a solution to our original equation.

And just like that, we've identified the single, true solution to our equation! The only valid solution for $\sqrt{x+66}=x+10$ is $\mathbf{x=-2}$. This emphasizes why verification is not just a suggestion, but a mandatory step for solving radical equations. Without it, we might have incorrectly concluded that $-17$ was also a solution, leading to an incorrect answer.

Why Extraneous Solutions Pop Up: The Nitty-Gritty Details

Okay, so we've seen an extraneous solution in action with $x=-17$. But why exactly do these imposters appear? It's not some arbitrary rule; there's a deep mathematical reason behind it, and understanding it will make you a much stronger problem-solver. The culprit, my friends, is that step where we squared both sides of the equation. Let's break down the logic.

Consider our original equation: $\sqrt{x+66}=x+10$. One of the fundamental properties of the square root symbol is that $\sqrt{A}$ (the principal square root) is always defined as the non-negative root of A. This means $\sqrt{x+66}$ must be greater than or equal to zero. Consequently, for the equation to hold, the right side, $x+10$, must also be greater than or equal to zero. This is a domain restriction implicitly imposed by the radical sign itself. So, any potential solution 'x' must satisfy $x+10 \ge 0$, or $x \ge -10$.

Now, when we squared both sides, we went from:

$\sqrt{x+66} = x+10 \quad \text{ (Equation A, implies } x+10 \ge 0 \text{)}$

to

$(\sqrt{x+66})^2 = (x+10)^2$

which simplified to:

$x+66 = (x+10)^2 \quad \text{ (Equation B)}$

Here's the critical point: Equation B is actually derived from two possibilities related to Equation A's structure:

1. The case where $\sqrt{x+66} = x+10$ (which implies $x+10 \ge 0$)
2. The case where $\sqrt{x+66} = -(x+10)$ (which also implies $x+10 \ge 0$ for $\sqrt{x+66}$ to be positive, but this would mean $-(x+10)$ would be positive, so $x+10$ would be negative. This is where the conflict arises!)

Let's rephrase this more clearly. When you square both sides of an equation, you're essentially performing an operation that doesn't distinguish between positive and negative values. If $A = B$, then $A^2 = B^2$. But it's also true that if $A = -B$, then $A^2 = (-B)^2 = B^2$. So, when we square $\sqrt{x+66}=x+10$, we essentially create an equation, $x+66=(x+10)^2$, that would be true not only if $\sqrt{x+66}=x+10$ but also if $\sqrt{x+66}=-(x+10)$.

Let's test our extraneous solution, $x=-17$, in this light:

• Original Equation: $\sqrt{-17+66} = -17+10 \implies \sqrt{49} = -7 \implies 7 = -7$ (False)

Now, consider the alternative form that squaring introduced: $\sqrt{x+66} = -(x+10)$.

• Alternative Form with $x=-17$: $\sqrt{-17+66} = -(-17+10) \implies \sqrt{49} = -(-7) \implies 7 = 7$ (True!)

See? The solution $x=-17$ is actually a solution to the equation $\sqrt{x+66}=-(x+10)$, not our original equation $\sqrt{x+66}=x+10$. Our squaring process introduced this "ghost" equation into our solution set, giving us results that weren't valid for the original problem. This is why checking is so paramount! It filters out the solutions that only work for the squared version of the equation, but not for the one you started with, especially considering the non-negative nature of the principal square root. So, the next time you square both sides of a radical equation, remember you're opening the door to these