Solve √(2x + 15) - √(x - 1) = 3: Find X Values
Hey guys! Let's dive into solving this radical equation. It looks a bit intimidating at first, but we'll break it down step by step to find all possible values of x that satisfy the equation √(2x + 15) - √(x - 1) = 3. We'll also make sure to verify our solutions, because, you know, it's always good to double-check your work!
Understanding the Problem
Before we jump into the algebraic manipulations, let's understand what this equation is telling us. We have two square root terms, and their difference is equal to 3. Remember, the square root function only deals with non-negative numbers, so we need to keep in mind that the expressions inside the square roots, 2x + 15 and x - 1, must be greater than or equal to zero. This will give us some initial constraints on the possible values of x.
Setting Up the Domain
First, let's consider the expression inside the first square root: 2x + 15. For this to be non-negative, we need to solve the inequality:
2x + 15 ≥ 0
Subtracting 15 from both sides gives:
2x ≥ -15
Dividing by 2, we get:
x ≥ -7.5
Now, let's look at the expression inside the second square root: x - 1. For this to be non-negative, we need:
x - 1 ≥ 0
Adding 1 to both sides gives:
x ≥ 1
Combining these two inequalities, we see that x must be greater than or equal to 1 for both square roots to be defined. So, our domain is x ≥ 1. This is crucial because any solution we find later must satisfy this condition. If we get a value of x less than 1, we know it's an extraneous solution.
Solving the Equation
Okay, now that we've got our domain sorted, let's tackle the equation itself. The best way to deal with radical equations is usually to isolate one of the square roots and then square both sides. This will eliminate one of the radicals. So, let's rewrite the equation as:
√(2x + 15) = √(x - 1) + 3
Now, we square both sides. Remember to square the entire right side, not just the individual terms. This means we'll need to use the formula (a + b)² = a² + 2ab + b²:
(√(2x + 15))² = (√(x - 1) + 3)²
This simplifies to:
2x + 15 = (x - 1) + 6√(x - 1) + 9
Notice that we still have a square root term, but we've managed to eliminate one of them. Let's simplify and isolate the remaining square root term. Combine the constants on the right side:
2x + 15 = x + 8 + 6√(x - 1)
Subtract x and 8 from both sides:
x + 7 = 6√(x - 1)
Now, we square both sides again to get rid of the remaining square root:
(x + 7)² = (6√(x - 1))²
Expanding the left side and squaring the right side, we get:
x² + 14x + 49 = 36(x - 1)
Distribute the 36 on the right side:
x² + 14x + 49 = 36x - 36
Now, we have a quadratic equation! Let's move all the terms to one side to set the equation equal to zero:
x² + 14x - 36x + 49 + 36 = 0
Combine like terms:
x² - 22x + 85 = 0
Solving the Quadratic Equation
We now have a quadratic equation in the standard form ax² + bx + c = 0. We can solve this using factoring, the quadratic formula, or completing the square. Let's try factoring first. We're looking for two numbers that multiply to 85 and add up to -22. Those numbers are -17 and -5. So, we can factor the quadratic as:
(x - 17)(x - 5) = 0
This gives us two potential solutions:
x - 17 = 0 => x = 17
x - 5 = 0 => x = 5
Verifying the Solutions
We've found two possible solutions, x = 17 and x = 5. But remember, we need to check if these solutions actually satisfy the original equation. This is super important because squaring both sides can sometimes introduce extraneous solutions (solutions that don't work in the original equation).
Checking x = 17
Let's plug x = 17 into the original equation:
√(2(17) + 15) - √(17 - 1) = 3
√(34 + 15) - √(16) = 3
√49 - √16 = 3
7 - 4 = 3
3 = 3
So, x = 17 is a valid solution! Awesome!
Checking x = 5
Now, let's plug in x = 5:
√(2(5) + 15) - √(5 - 1) = 3
√(10 + 15) - √4 = 3
√25 - √4 = 3
5 - 2 = 3
3 = 3
x = 5 is also a valid solution! Double awesome!
Final Answer
Therefore, the possible values of x that satisfy the equation √(2x + 15) - √(x - 1) = 3 are x = 5 and x = 17. We found the solutions by isolating the square roots, squaring both sides (twice!), solving the resulting quadratic equation, and most importantly, verifying our solutions in the original equation. This is a classic example of how to solve radical equations, guys. Keep practicing, and you'll become a pro in no time!
