Solving 4 + √(5x + 66) = X + 10 A Step-by-Step Guide
Hey guys! Today, we're diving deep into the world of algebra to tackle an interesting equation: 4 + √(5x + 66) = x + 10. This isn't just any equation; it involves a square root, which means we need to be extra careful with our steps to avoid any pitfalls. We'll break down the solution step-by-step, making sure you understand the why behind each move. So, grab your pencils, and let's get started!
Isolating the Square Root
In solving equations involving radicals, the first crucial step is to isolate the square root term. This means we want to get the √(5x + 66) part all by itself on one side of the equation. To do this, we need to get rid of that pesky 4 that's hanging out on the left side.
How do we do that? Simple! We subtract 4 from both sides of the equation. Remember, whatever we do to one side, we must do to the other to keep the equation balanced. This is a fundamental principle in algebra, and it's super important to remember. So, let's do it:
4 + √(5x + 66) - 4 = x + 10 - 4
This simplifies to:
√(5x + 66) = x + 6
Awesome! Now we have the square root term isolated. This sets us up for the next big step: squaring both sides.
Squaring Both Sides
Now that we've isolated the square root, it's time to get rid of it altogether. The way we do this is by squaring both sides of the equation. Think of it this way: the opposite of taking a square root is squaring a number. So, squaring the square root will effectively cancel it out. But again, a crucial reminder: what we do to one side, we must do to the other.
So, let's square both sides:
(√(5x + 66))^2 = (x + 6)^2
On the left side, the square root and the square cancel each other out, leaving us with:
5x + 66
The right side is a little trickier. We're squaring the binomial (x + 6), which means we need to multiply it by itself:
(x + 6)^2 = (x + 6)(x + 6)
We can use the FOIL method (First, Outer, Inner, Last) to expand this:
- First: x * x = x²
- Outer: x * 6 = 6x
- Inner: 6 * x = 6x
- Last: 6 * 6 = 36
Adding these together, we get:
x² + 6x + 6x + 36 = x² + 12x + 36
So, our equation now looks like this:
5x + 66 = x² + 12x + 36
We've successfully eliminated the square root! Now we're dealing with a quadratic equation, which we know how to solve.
Rearranging into a Quadratic Equation
Okay, now we've got a quadratic equation staring us in the face: 5x + 66 = x² + 12x + 36. To solve this, we need to get all the terms on one side and set the equation equal to zero. This will put it in the standard quadratic form: ax² + bx + c = 0.
To do this, let's subtract 5x and 66 from both sides of the equation. This will move all the terms to the right side, leaving zero on the left:
5x + 66 - 5x - 66 = x² + 12x + 36 - 5x - 66
Simplifying, we get:
0 = x² + 7x - 30
Great! Now our quadratic equation is in standard form: x² + 7x - 30 = 0. This is perfect for factoring, which is our next step.
Factoring the Quadratic Equation
The next key step is to factor the quadratic equation x² + 7x - 30 = 0. Factoring involves breaking down the quadratic expression into two binomials. We're looking for two numbers that:
- Multiply to give us the constant term (-30)
- Add up to give us the coefficient of the x term (7)
Let's think about the factors of -30. We need one positive and one negative number since the product is negative. After a little brainstorming, we can see that 10 and -3 fit the bill:
- 10 * -3 = -30
- 10 + (-3) = 7
Perfect! So, we can factor the quadratic equation as follows:
(x + 10)(x - 3) = 0
This means that either (x + 10) = 0 or (x - 3) = 0. This is a crucial step because it allows us to find the possible values of x.
Solving for x
Now that we've factored the quadratic equation, we're in the home stretch. We have two factors, and for the product of these factors to be zero, at least one of them must be zero. This gives us two separate equations to solve:
- x + 10 = 0
- x - 3 = 0
Let's solve each one individually.
For the first equation, x + 10 = 0, we subtract 10 from both sides:
x + 10 - 10 = 0 - 10
This gives us:
x = -10
So, one possible solution is x = -10.
For the second equation, x - 3 = 0, we add 3 to both sides:
x - 3 + 3 = 0 + 3
This gives us:
x = 3
So, another possible solution is x = 3.
We have two potential solutions: x = -10 and x = 3. But hold on! We're not done yet. Because we squared both sides of the equation earlier, we need to check our solutions to make sure they actually work. This is super important because squaring can sometimes introduce extraneous solutions – solutions that don't satisfy the original equation.
Checking for Extraneous Solutions
This is the most crucial step in solving radical equations. Because we squared both sides earlier, it's possible that we introduced extraneous solutions – values that satisfy the transformed equation but not the original one. We need to plug each potential solution back into the original equation to see if it holds true.
Our original equation was:
4 + √(5x + 66) = x + 10
Let's start by checking x = -10:
4 + √(5(-10) + 66) = -10 + 10
4 + √(-50 + 66) = 0
4 + √16 = 0
4 + 4 = 0
8 = 0
This is clearly not true, so x = -10 is an extraneous solution. It doesn't work in the original equation.
Now, let's check x = 3:
4 + √(5(3) + 66) = 3 + 10
4 + √(15 + 66) = 13
4 + √81 = 13
4 + 9 = 13
13 = 13
This is true, so x = 3 is a valid solution.
The Final Answer
After all that work, we've arrived at the final answer. We started with the equation 4 + √(5x + 66) = x + 10, and after carefully isolating the square root, squaring both sides, solving the resulting quadratic equation, and – most importantly – checking for extraneous solutions, we found that only one solution holds true.
Therefore, the solution to the equation is:
x = 3
So, the correct answer is B. x = 3.
Remember, guys, solving equations with square roots can be a bit tricky, but by following these steps carefully, you can conquer any radical equation that comes your way. Keep practicing, and you'll become a master of algebra in no time!
