Solve $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$

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Hey guys! Today, we're diving deep into the world of radical equations. We're going to break down how to solve the equation $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$. This looks a bit intimidating at first, but don't worry, we'll tackle it together step by step. We'll focus on making sure you not only get the right answer but also understand the why behind each step. So, grab your thinking caps, and let's get started!

Understanding the Domain

Before we even think about manipulating the equation, it's super important to figure out the domain of $x$. Remember, we can't take the square root of a negative number (at least, not in the realm of real numbers!). So, we need to make sure everything under the square roots is non-negative. This is a crucial first step in solving any radical equation, as it helps us avoid extraneous solutions later on. Extraneous solutions are basically answers that pop up during the solving process but don't actually work in the original equation.

Let's break down the domain restrictions:

1. For $\sqrt{x+3}$, we need $x+3 \geq 0$, which means $x \geq -3$.
2. For $\sqrt{5-x}$, we need $5-x \geq 0$, which means $x \leq 5$.
3. For $\sqrt{15+2x-x^2}$, we need $15+2x-x^2 \geq 0$. This is a quadratic, so let's factor it. We can rewrite it as $-(x^2 - 2x - 15) \geq 0$, which factors into $-(x-5)(x+3) \geq 0$. Multiplying both sides by -1 (and flipping the inequality sign!) gives us $(x-5)(x+3) \leq 0$. This inequality holds true when $-3 \leq x \leq 5$.

Combining all these restrictions, we see that the domain of $x$ is indeed $[-3, 5]$. This means any solution we find must fall within this interval. If it doesn't, we know we've made a mistake somewhere or encountered an extraneous solution. Keep this domain in mind as we move forward!

Simplifying the Equation

Okay, now that we've got the domain nailed down, let's get to simplifying the equation. The key here is to notice that $15+2x-x^2$ is actually the product of $(x+3)$ and $(5-x)$. This is a huge observation because it allows us to rewrite the equation in a much cleaner way. Spotting patterns like this is a crucial skill in algebra, and it often makes seemingly complex problems much more manageable.

So, we can rewrite $\sqrt{15+2x-x^2}$ as $\sqrt{(x+3)(5-x)}$. Now, let's substitute this back into our original equation:

$\sqrt{x+3} + \sqrt{5-x} - 2\sqrt{(x+3)(5-x)} = -4$

This looks much better already! We've got a common structure emerging, which is always a good sign. Think of it like finding a common denominator when adding fractions – it sets us up for the next simplification step. The goal here is to make the equation easier to manipulate, and recognizing this factorization is a big step in that direction.

Introducing Substitution

Now for the magic trick – substitution! This is a powerful technique in algebra that allows us to replace complex expressions with simpler variables, making the equation look much less scary. In this case, we're going to let $a = \sqrt{x+3}$ and $b = \sqrt{5-x}$. This might seem a bit abstract at first, but you'll see how it cleans things up nicely.

Notice that with this substitution, our equation transforms into:

$a + b - 2ab = -4$

Wow, that's a lot simpler, right? We've gone from a radical equation with square roots all over the place to a relatively simple algebraic equation in terms of $a$ and $b$. But we're not done yet! We have two variables, $a$ and $b$, and only one equation. To solve for them, we need another equation that relates $a$ and $b$.

Think about how $a$ and $b$ are defined. We have $a = \sqrt{x+3}$ and $b = \sqrt{5-x}$. What happens if we square both of these equations? We get $a^2 = x+3$ and $b^2 = 5-x$. Now, what if we add these two equations together? Something very convenient happens!

$a^2 + b^2 = (x+3) + (5-x) = 8$

Look at that! The $x$ terms cancel out, leaving us with a simple equation relating $a^2$ and $b^2$. This is our second equation, and now we have a system of equations:

1. $a + b - 2ab = -4$
2. $a^2 + b^2 = 8$

We've successfully transformed our original radical equation into a system of two algebraic equations. This is a common strategy in solving complex problems – break them down into smaller, more manageable parts. Now, let's tackle this system!

Solving the System of Equations

Okay, guys, we've got our system of equations: $a + b - 2ab = -4$ and $a^2 + b^2 = 8$. Now, the question is, how do we solve this? There are a few different approaches we could take, but a common strategy for systems of equations is to try to isolate one variable in one equation and substitute it into the other. However, in this case, directly isolating $a$ or $b$ in the first equation might lead to some messy algebra. Instead, let's try a clever trick that leverages a common algebraic identity.

Remember the identity $(a+b)^2 = a^2 + 2ab + b^2$? We already know $a^2 + b^2 = 8$, so if we can find $(a+b)$, we can then find $ab$, and vice versa. This is a powerful technique for solving systems of equations where squares and products are involved. Let's see how we can use it here.

Let's rearrange our first equation, $a + b - 2ab = -4$, to isolate the $2ab$ term:

$2ab = a + b + 4$

Now, let's consider $(a+b)^2$. Using the identity, we have:

$(a+b)^2 = a^2 + 2ab + b^2$

We know $a^2 + b^2 = 8$ and we have an expression for $2ab$, so let's substitute those in:

$(a+b)^2 = 8 + (a + b + 4)$

This simplifies to:

$(a+b)^2 = a + b + 12$

Now, this is looking good! We have an equation involving only $(a+b)$. Let's make another substitution to make it even clearer. Let $S = a + b$. Then our equation becomes:

$S^2 = S + 12$

This is a simple quadratic equation! We can rearrange it to get:

$S^2 - S - 12 = 0$

This factors nicely into:

$(S-4)(S+3) = 0$

So, we have two possible values for $S = a + b$: $S = 4$ or $S = -3$.

Now, let's consider each case separately. This is a crucial step – we need to explore all possibilities to ensure we find all solutions.

Case 1: $a + b = 4$

If $a + b = 4$, then we can substitute this back into our equation $2ab = a + b + 4$ to get:

$2ab = 4 + 4 = 8$

So, $ab = 4$.

Now we have $a + b = 4$ and $ab = 4$. This reminds us of another important concept: Vieta's formulas! Vieta's formulas relate the coefficients of a polynomial to the sums and products of its roots. In this case, we can think of $a$ and $b$ as the roots of a quadratic equation.

Specifically, $a$ and $b$ are the roots of the quadratic equation:

$t^2 - (a+b)t + ab = 0$

Substituting in our values for $a+b$ and $ab$, we get:

$t^2 - 4t + 4 = 0$

This factors into:

$(t-2)^2 = 0$

So, $t = 2$ is a repeated root. This means $a = b = 2$.

Case 2: $a + b = -3$

If $a + b = -3$, then substituting into $2ab = a + b + 4$ gives us:

$2ab = -3 + 4 = 1$

So, $ab = \frac{1}{2}$.

Now we have $a + b = -3$ and $ab = \frac{1}{2}$. Again, using Vieta's formulas, $a$ and $b$ are the roots of the quadratic equation:

$t^2 - (a+b)t + ab = 0$

Substituting, we get:

$t^2 + 3t + \frac{1}{2} = 0$

To get rid of the fraction, let's multiply the entire equation by 2:

$2t^2 + 6t + 1 = 0$

This quadratic doesn't factor easily, so we'll use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4(2)(1)}}{2(2)} = \frac{-6 \pm \sqrt{28}}{4} = \frac{-6 \pm 2\sqrt{7}}{4} = \frac{-3 \pm \sqrt{7}}{2}$

So, in this case, we have two possible pairs for $(a, b)$: $\left( \frac{-3 + \sqrt{7}}{2}, \frac{-3 - \sqrt{7}}{2} \right)$ and $\left( \frac{-3 - \sqrt{7}}{2}, \frac{-3 + \sqrt{7}}{2} \right)$.

Back-Substitution and Finding x

Alright, we've found our values for $a$ and $b$, but remember, we're trying to solve for $x$! This means we need to back-substitute and use our original definitions of $a$ and $b$ to find the corresponding values of $x$.

Let's start with Case 1, where $a = b = 2$. Recall that $a = \sqrt{x+3}$ and $b = \sqrt{5-x}$. So, we have:

$\sqrt{x+3} = 2$ and $\sqrt{5-x} = 2$

Squaring both sides of each equation, we get:

$x + 3 = 4$ and $5 - x = 4$

Both of these equations give us $x = 1$. This looks like a promising solution, but remember, we need to check if it's in our domain $[-3, 5]$. Since $1$ is within this domain, it's a potential solution. We'll verify it later.

Now let's move on to Case 2, where we have $a = \frac{-3 + \sqrt{7}}{2}$ and $b = \frac{-3 - \sqrt{7}}{2}$ (or vice versa). Since square roots cannot be negative we can disregard this case, since $\frac{-3 - \sqrt{7}}{2}$ and $\frac{-3 + \sqrt{7}}{2}$ are negative. Note that $\sqrt{7}$ is slightly less than 3, making $\frac{-3 + \sqrt{7}}{2}$ negative and $\frac{-3 - \sqrt{7}}{2}$ is certainly negative.

Verification and Final Solution

We've found a potential solution: $x = 1$. But we're not done until we verify it in the original equation. Plugging $x = 1$ into $\sqrt{x+3} + \sqrt{5-x} - 2\sqrt{15+2x-x^2} = -4$, we get:

$\sqrt{1+3} + \sqrt{5-1} - 2\sqrt{15+2(1)-1^2} = \sqrt{4} + \sqrt{4} - 2\sqrt{16} = 2 + 2 - 2(4) = 4 - 8 = -4$

It works! So, $x = 1$ is indeed a solution.

Therefore, the solution to the equation $\sqrt{x+3} + \sqrt{5-x} - 2\sqrt{15+2x-x^2} = -4$ is $x = 1$.

We made it! We successfully navigated this radical equation by remembering the importance of domains, recognizing key factorizations, using substitution to simplify the problem, solving a system of equations, and finally, verifying our solution. Solving these kinds of problems isn't just about getting the right answer; it's about developing a toolbox of problem-solving techniques that you can apply to all sorts of math challenges. Keep practicing, and you'll become a radical equation master in no time!