Solve $\sqrt{x^2+49}=x+5$?

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Hey math whizzes and puzzle solvers! Today, we're diving headfirst into a super common type of problem you'll bump into in algebra: solving radical equations. Specifically, we're tackling the equation $\sqrt{x^2+49}=x+5$. Now, I know what some of you might be thinking – "Square roots? Equations? This sounds complicated!" But trust me, guys, with a few key steps and a bit of careful thinking, this kind of problem becomes totally manageable. We're going to break down exactly how to solve this, why it's important to check your answers, and what pitfalls to watch out for. So, grab your notebooks, maybe a snack, and let's get ready to conquer this equation together. We'll explore the logic behind isolating the radical, squaring both sides, and most importantly, the crucial step of verifying our solutions to ensure we haven't accidentally introduced any extraneous roots. By the end of this, you'll feel way more confident tackling similar problems, and you'll understand the 'why' behind each step, not just the 'how'. We're aiming to make this as clear and straightforward as possible, so even if you're new to this stuff, you'll be able to follow along and learn a ton. Let's get this mathematical party started!

The Core Challenge: Dealing with the Square Root

The main hurdle in solving $\sqrt{x^2+49}=x+5$ is that pesky square root symbol. Our goal in solving any equation is usually to get the variable (in this case, 'x') all by itself on one side. But that square root is acting like a stubborn bodyguard, preventing us from doing that easily. So, the first big strategy we need to employ is to isolate the radical. This means getting the term with the square root completely on one side of the equals sign, with nothing else added or subtracted from it. In our equation, $\sqrt{x^2+49}=x+5$, the radical is already isolated on the left side, which is convenient! If it wasn't, say we had something like $\sqrt{x^2+49} - 3 = x + 2$, our first step would be to add 3 to both sides to get $\sqrt{x^2+49} = x+5$. So, because our equation is already set up nicely, we can move straight to the next, and arguably the most powerful, step: eliminating the square root by squaring both sides. Remember, the inverse operation of taking a square root is squaring. Whatever you do to one side of an equation, you must do to the other to maintain balance. So, we're going to square the entire left side ($\left(\sqrt{x^2+49}\right)^2$) and the entire right side ($(x+5)^2$). This is where the magic happens and the square root symbol disappears, leaving us with a much simpler equation to solve. It’s crucial to remember that squaring both sides can sometimes introduce extraneous solutions, which are answers that work in the squared equation but not in the original. We'll talk more about why this happens and how to catch them later, but for now, just know that this is a critical step that transforms the problem into a more familiar algebraic form. So, the game plan is: isolate the radical (already done here!), then square both sides to get rid of it.

Squaring Both Sides: The Transformation

Alright, guys, we've isolated our radical term, $\sqrt{x^2+49}$, and now it's time for the big move: squaring both sides of the equation. This is the key to transforming our radical equation into a polynomial equation, which we know how to solve! Let's take our original equation: $\sqrt{x^2+49} = x+5$. We're going to apply the square operation to both sides:

$\left(\sqrt{x^2+49}\right)^2 = (x+5)^2$

On the left side, the square and the square root cancel each other out perfectly, leaving us with just the expression inside the radical:

$x^2+49$

Now, for the right side, we need to be super careful. We can't just square the 'x' and square the '5' individually. We have to expand $(x+5)^2$. Remember the rule for squaring a binomial? It's $(a+b)^2 = a^2 + 2ab + b^2$. In our case, $a=x$ and $b=5$. So, $(x+5)^2$ becomes:

$x^2 + 2(x)(5) + 5^2$

Which simplifies to:

$x^2 + 10x + 25$

Now, let's put both sides back together. Our equation has transformed from $\sqrt{x^2+49}=x+5$ into:

$x^2 + 49 = x^2 + 10x + 25$

See? No more square roots! This looks much more like a standard algebra problem. The next step, naturally, is to simplify this new equation and solve for 'x'. We'll want to gather all the 'x' terms on one side and the constant terms on the other. Notice that we have an $x^2$ term on both sides. This is great news because we can subtract $x^2$ from both sides, which will cancel them out, leaving us with a linear equation (an equation where the highest power of x is 1), which is usually the easiest type to solve. This squaring step is fundamental to solving radical equations, but always remember the potential for extraneous solutions, which we'll address next!

Simplifying and Solving: The Linear Equation Emerges

Okay, math crew, we've squared both sides of $\sqrt{x^2+49}=x+5$ and arrived at $x^2 + 49 = x^2 + 10x + 25$. Now, the mission is to simplify this equation and find the value(s) of 'x'. As we spotted in the last step, we have $x^2$ on both the left and the right sides. This is fantastic because it means our equation is about to become much simpler. We can eliminate the $x^2$ terms by performing the same operation on both sides of the equation. Let's subtract $x^2$ from both sides:

$(x^2 + 49) - x^2 = (x^2 + 10x + 25) - x^2$

This simplifies beautifully to:

$49 = 10x + 25$

Boom! Just like that, our quadratic equation (with $x^2$) has transformed into a linear equation (with just 'x'). This is the kind of equation we're super comfortable solving. Our next goal is to isolate the 'x' term. To do this, we first want to get all the constant terms on the side opposite the 'x' term. Since $10x$ is on the right, let's move the '25' over to the left. We do this by subtracting 25 from both sides:

$49 - 25 = 10x + 25 - 25$

This gives us:

$24 = 10x$

We're almost there! The 'x' is being multiplied by 10. To get 'x' by itself, we need to perform the inverse operation, which is division. Let's divide both sides by 10:

rac{24}{10} = rac{10x}{10}

And that leaves us with:

x = rac{24}{10}

Now, we should always simplify our fractions if possible. Both 24 and 10 are divisible by 2:

x = rac{12}{5}

So, our potential solution is x = rac{12}{5}. But hold on! Remember what we talked about? Squaring both sides can sometimes create fake solutions. We must check this answer in the original equation to make sure it's the real deal. This checking step is non-negotiable in solving radical equations. Let's move on to that crucial verification process.

The Crucial Check: Are There Extraneous Solutions?

Alright, team, we've worked hard and found a potential solution, x = rac{12}{5}, for the equation $\sqrt{x^2+49}=x+5$. But here's the most important step when dealing with radical equations, especially after squaring both sides: we have to check our answer. Why? Because squaring both sides can sometimes introduce extraneous solutions. An extraneous solution is a value that satisfies the manipulated equation (the one after squaring) but does not satisfy the original equation. It's like finding a key that fits a modified lock but not the original one. So, let's plug our potential solution, x = rac{12}{5}, back into the original equation: $\sqrt{x^2+49}=x+5$.

Let's calculate the left side (LS) first:

LS = $\sqrt{\left(\frac{12}{5}\right)^2 + 49}$

LS = $\sqrt{\frac{144}{25} + 49}$

To add these, we need a common denominator. $49 = \frac{49 \times 25}{25} = \frac{1225}{25}$.

LS = $\sqrt{\frac{144}{25} + \frac{1225}{25}}$

LS = $\sqrt{\frac{144 + 1225}{25}}$

LS = $\sqrt{\frac{1369}{25}}$

Now, we need to find the square root of 1369. If you try it, you'll find that $37^2 = 1369$. So:

LS = $\frac{\sqrt{1369}}{\sqrt{25}}$

LS = $\frac{37}{5}$

Great! The left side evaluates to $\frac{37}{5}$. Now, let's calculate the right side (RS) using $x = \frac{12}{5}$:

RS = $x + 5$

RS = $\frac{12}{5} + 5$

Again, find a common denominator. $5 = \frac{5 \times 5}{5} = \frac{25}{5}$.

RS = $\frac{12}{5} + \frac{25}{5}$

RS = $\frac{12 + 25}{5}$

RS = $\frac{37}{5}$

Look at that! The left side ($\frac{37}{5}$) is equal to the right side ($\frac{37}{5}$). Since both sides match when we plug in $x = \frac{12}{5}$, this means our solution is valid and not extraneous. We found the one true answer!

Final Answer and Why Other Options Don't Work

So, after all that hard work, we've confirmed that $x = \frac{12}{5}$ is indeed the solution to the equation $\sqrt{x^2+49}=x+5$. We meticulously followed the steps: isolating the radical, squaring both sides, simplifying the resulting equation, and – crucially – checking our potential solution in the original equation. The check showed that when $x = \frac{12}{5}$, the left side equals the right side, meaning it's a valid solution.

Now, let's quickly look at the other options provided to understand why they aren't the correct answer for this specific problem:

• $x = -\frac{12}{5}$: If we were to plug this value back into the original equation, we would find that the left side ($\sqrt{(-12/5)^2 + 49} = \sqrt{144/25 + 49} = \sqrt{1369/25} = 37/5$) does not equal the right side ($-12/5 + 5 = -12/5 + 25/5 = 13/5$). So, this is an extraneous solution.
• $x = -6$ or $x = -3$: These values would likely arise from errors in algebraic manipulation, particularly during the squaring or simplification stages. Plugging either of these into the original equation would quickly show they don't satisfy it.
• No solution: This option would be correct if, after checking, our potential solution ($x=12/5$ in this case) failed the verification step, or if the process led to a contradiction (like $0=5$). Since $x=12/5$ did work, there is a solution.

Therefore, the only correct solution is $x = \frac{12}{5}$. It's a great reminder that in algebra, especially with equations involving roots or other tricky functions, the final check is your best friend for accuracy. Keep practicing these steps, guys, and you'll become equation-solving ninjas in no time!