Solving For $w$: A Real Number Equation

Hey guys! Today, we're going to dive into solving an equation where we need to find the value of $w$, given that $w$ is a real number. The equation we're tackling is $11 + \sqrt{w-7} = 5$. This type of problem involves a square root, so we need to be extra careful with our steps to make sure we don't introduce any extraneous solutions. Let's break it down and solve it together!

Understanding the Equation

Before we jump into the algebraic manipulations, let's take a moment to understand what the equation is telling us. We have a square root term, $\sqrt{w-7}$, which means that $w-7$ must be greater than or equal to zero because we're dealing with real numbers. Otherwise, we'd be venturing into the realm of imaginary numbers, which isn't what we want in this case. The equation states that 11 plus this square root is equal to 5. Intuitively, we can already see that the square root term must be negative since 5 is less than 11. This gives us a clue that we might run into some issues, but let's proceed systematically.

Also, remember that the square root function, when referring to the principal square root, always yields a non-negative value. This is a crucial point that will help us determine if our solution is valid. Essentially, $\sqrt{x}$ for any real number $x$ (where $x \geq 0$) will always be greater than or equal to 0. This understanding is key to solving equations involving square roots, as it helps us identify potential extraneous solutions.

By keeping these concepts in mind, we can approach the problem with a clearer understanding of what to expect and how to interpret our results. Now, let's proceed with the steps to isolate $w$ and find its value.

Step-by-Step Solution

Here’s how we can solve for $w$:

1. Isolate the Square Root Term: Our first goal is to isolate the square root on one side of the equation. To do this, we subtract 11 from both sides:

   $\sqrt{w-7} = 5 - 11$ $\sqrt{w-7} = -6$

2. Analyze the Result: Now, we have $\sqrt{w-7} = -6$. Remember that the square root of a real number cannot be negative. The principal square root is always non-negative. Therefore, $\sqrt{w-7}$ must be greater than or equal to 0. Since we have $\sqrt{w-7} = -6$, which is a negative number, there is no real number $w$ that can satisfy this equation. This is because the square root function, by definition, returns the non-negative root.

3. Conclusion: Because the square root of a real number cannot be negative, there is no solution to this equation.

Why Squaring Both Sides Doesn't Work

You might be wondering, "What if I squared both sides?" Let's explore that scenario:

Starting from $\sqrt{w-7} = -6$, if we square both sides, we get:

$(\sqrt{w-7})^2 = (-6)^2$ $w - 7 = 36$ $w = 36 + 7$ $w = 43$

Now, let’s check if $w = 43$ is a valid solution by substituting it back into the original equation:

$11 + \sqrt{43 - 7} = 5$ $11 + \sqrt{36} = 5$ $11 + 6 = 5$ $17 = 5$

This is clearly false! So, $w = 43$ is an extraneous solution. This happened because squaring both sides of an equation can introduce solutions that don't actually satisfy the original equation, especially when dealing with square roots or absolute values. Squaring both sides can turn a false statement into a true one. For example, $-2 \neq 2$, but $(-2)^2 = 2^2 = 4$.

The Importance of Checking Solutions

This example highlights the critical importance of checking your solutions when solving equations, especially those involving radicals. Whenever you perform an operation that could introduce extraneous solutions (like squaring both sides), you must substitute your solutions back into the original equation to verify that they are valid. If a solution doesn't satisfy the original equation, it's an extraneous solution and must be discarded.

In our case, squaring both sides led us to $w = 43$, but substituting this value back into the original equation showed that it was not a valid solution. This reinforces the idea that checking solutions is a crucial step in the problem-solving process.

Domain Considerations

Another important aspect to consider when dealing with square roots is the domain of the variable. In the original equation, $11 + \sqrt{w-7} = 5$, the expression inside the square root, $w-7$, must be greater than or equal to zero. This gives us the inequality:

$w - 7 \geq 0$ $w \geq 7$

This means that any solution we find for $w$ must be greater than or equal to 7. If we find a solution that doesn't satisfy this condition, it cannot be a valid solution. In our case, even if we had found a numerical value for $w$ that seemed to work after squaring both sides, we would still need to check if it satisfies the condition $w \geq 7$.

No Solution

So, to wrap it up, the equation $11 + \sqrt{w-7} = 5$ has no solution in the real number system. The key takeaway is that the square root of a real number cannot be negative, and checking for extraneous solutions is crucial when solving equations with radicals. Keep practicing, and you'll become a pro at solving these types of problems!