Solving The Equation √(-2x - 5) - 4 = X Step-by-Step
In this article, we will explore how to solve the equation √(-2x - 5) - 4 = x. This equation involves a square root, which means we need to be careful about extraneous solutions. We'll break down the steps, discuss potential pitfalls, and arrive at the correct solution. Understanding how to solve equations with square roots is crucial in algebra and many areas of mathematics. Let's dive into the process and find the values of x that satisfy this equation.
Isolating the Square Root
The first step in solving an equation with a square root is to isolate the square root term. This means getting the term with the square root by itself on one side of the equation. In our case, the equation is:
√(-2x - 5) - 4 = x
To isolate the square root, we need to add 4 to both sides of the equation:
√(-2x - 5) - 4 + 4 = x + 4
This simplifies to:
√(-2x - 5) = x + 4
Now we have the square root term isolated, which sets us up for the next step. Isolating the square root is a critical move because it allows us to eliminate the square root by squaring both sides, which we will do in the next section. Remember, keeping the equation balanced is key to maintaining its validity. This isolation step is the foundation for successfully solving radical equations.
Squaring Both Sides
Now that we have isolated the square root term, the next step is to eliminate the square root. We achieve this by squaring both sides of the equation. This is a valid algebraic operation as long as we apply it to the entire side of the equation. Starting with our isolated equation:
√(-2x - 5) = x + 4
We square both sides:
(√(-2x - 5))² = (x + 4)²
On the left side, the square root and the square cancel each other out, leaving:
-2x - 5
On the right side, we need to expand the square of the binomial (x + 4). Remember that (x + 4)² means (x + 4)(x + 4). Using the FOIL method (First, Outer, Inner, Last) or the distributive property, we get:
(x + 4)(x + 4) = x² + 4x + 4x + 16 = x² + 8x + 16
So our equation now looks like this:
-2x - 5 = x² + 8x + 16
Squaring both sides is a powerful technique, but it also introduces a potential issue: extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (the one after squaring) but do not satisfy the original equation. This is why we must always check our solutions in the original equation to ensure they are valid.
Rearranging into a Quadratic Equation
After squaring both sides, we have the equation:
-2x - 5 = x² + 8x + 16
This equation is a quadratic equation because it contains a term with x². To solve a quadratic equation, we typically want to rearrange it into the standard form:
ax² + bx + c = 0
To do this, we need to move all the terms to one side of the equation, leaving zero on the other side. We can add 2x and 5 to both sides:
-2x - 5 + 2x + 5 = x² + 8x + 16 + 2x + 5
This simplifies to:
0 = x² + 10x + 21
Now we have a quadratic equation in standard form: x² + 10x + 21 = 0. The next step is to solve this quadratic equation, which can be done by factoring, using the quadratic formula, or completing the square. In this case, factoring is the most straightforward method. Rearranging the equation into this standard form is crucial for applying these solution techniques effectively.
Solving the Quadratic Equation by Factoring
Now we have the quadratic equation:
x² + 10x + 21 = 0
To solve this equation by factoring, we need to find two numbers that multiply to 21 (the constant term) and add up to 10 (the coefficient of the x term). Let's think about the factors of 21:
- 1 and 21
- 3 and 7
The pair 3 and 7 add up to 10, so these are the numbers we need. We can rewrite the quadratic equation as:
(x + 3)(x + 7) = 0
Now, for the product of two factors to be zero, at least one of the factors must be zero. This gives us two possible equations:
x + 3 = 0 or x + 7 = 0
Solving these equations for x, we get:
x = -3 or x = -7
So, we have two potential solutions: x = -3 and x = -7. However, we need to remember the crucial step of checking for extraneous solutions because we squared both sides of the equation earlier. Factoring is an efficient method for solving quadratic equations, but the validity of the solutions must always be verified in the original equation, especially when dealing with radical equations.
Checking for Extraneous Solutions
We found two potential solutions to our equation: x = -3 and x = -7. Now we must check each of these solutions in the original equation to ensure they are valid and not extraneous. The original equation was:
√(-2x - 5) - 4 = x
Let's first check x = -3:
√(-2(-3) - 5) - 4 = -3
√(+6 - 5) - 4 = -3
√(1) - 4 = -3
1 - 4 = -3
-3 = -3
This is true, so x = -3 is a valid solution.
Now let's check x = -7:
√(-2(-7) - 5) - 4 = -7
√(14 - 5) - 4 = -7
√(9) - 4 = -7
3 - 4 = -7
-1 = -7
This is false, so x = -7 is an extraneous solution. Checking for extraneous solutions is absolutely vital when solving equations with square roots. The squaring operation can introduce solutions that don't actually satisfy the original equation. We've seen that x = -7 does not work, while x = -3 does. This highlights the importance of this verification step.
The Final Solution
After solving the equation √(-2x - 5) - 4 = x, we found two potential solutions: x = -3 and x = -7. However, after checking for extraneous solutions, we determined that x = -7 is not a valid solution because it does not satisfy the original equation. The only solution that works is x = -3.
Therefore, the solution to the equation √(-2x - 5) - 4 = x is x = -3.
This process demonstrates the importance of careful algebraic manipulation and the necessity of verifying solutions when dealing with radical equations. The correct answer is C. -3. Understanding these steps is key to solving similar equations confidently and accurately. The journey from isolating the square root to verifying the solutions showcases the intricacies of solving equations involving radicals, and mastering this process is a valuable skill in mathematics.
