Solving The Equation: S = 4 + √(s + 2)

Hey guys! Today, let's dive into solving a cool little equation: s = 4 + √(s + 2). This equation involves a square root, so we'll need to be a bit careful with our steps to make sure we arrive at the correct solution. We'll break it down step by step, making sure everyone can follow along. So, grab your thinking caps, and let's get started!

Understanding the Equation

Before we jump into the solution, let's first understand what we're dealing with. Our equation is s = 4 + √(s + 2). The key here is the square root, √(s + 2). Remember, the square root of a number is a value that, when multiplied by itself, gives you the original number. For example, the square root of 9 is 3 because 3 * 3 = 9. Also, it's important to remember that the value inside the square root (the radicand, which is s + 2 in our case) must be non-negative because we can't take the square root of a negative number and get a real number result. This gives us our first important consideration: s + 2 ≥ 0, which means s ≥ -2. This constraint will help us check if our final solutions are valid.

In this equation, we're trying to find the value(s) of 's' that make the equation true. This involves isolating 's' and getting rid of the square root. The general strategy for solving equations with square roots is to isolate the square root term and then square both sides of the equation. Squaring both sides gets rid of the square root, but it can also introduce extraneous solutions, which are solutions that we find algebraically but don't actually satisfy the original equation. That's why we need to check our answers at the end.

Why is it important to check for extraneous solutions? When we square both sides of an equation, we're essentially saying that if a = b, then a² = b². This is true, but the reverse is not necessarily true. If a² = b², it could be that a = b or a = -b. Think about it: both 3² and (-3)² equal 9. So, squaring can introduce a negative possibility that wasn't there originally. That's why we must plug our potential solutions back into the original equation to see if they work.

So, with that understanding in mind, let’s proceed with the steps to solve this equation. We will carefully isolate the square root, square both sides, solve the resulting equation, and most importantly, verify our solutions to eliminate any extraneous ones. This methodical approach will ensure we find the correct solution(s) for 's'.

Step-by-Step Solution

Alright, let's jump into the step-by-step solution of the equation s = 4 + √(s + 2). We'll take it one step at a time to make sure we're crystal clear on each part of the process. Remember, our goal is to isolate 's', but first, we need to get rid of that pesky square root!

Step 1: Isolate the Square Root

The first thing we need to do is isolate the square root term. This means we want to get the term √(s + 2) all by itself on one side of the equation. Currently, we have s = 4 + √(s + 2). To isolate the square root, we need to subtract 4 from both sides of the equation. This gives us:

s - 4 = √(s + 2)

Now, the square root term is isolated on the right side, which is exactly what we wanted. This sets us up perfectly for the next step, which involves getting rid of the square root altogether.

Step 2: Square Both Sides

To eliminate the square root, we'll square both sides of the equation. Remember, whatever we do to one side, we must do to the other to keep the equation balanced. So, we'll square both (s - 4) and √(s + 2). This looks like:

(s - 4)² = (√(s + 2))²

On the right side, the square and the square root cancel each other out, leaving us with just (s + 2). On the left side, we need to expand (s - 4)². Recall that (a - b)² = a² - 2ab + b². Applying this to our equation, we get:

s² - 8s + 16 = s + 2

Now we have a quadratic equation, which is great because we know how to solve those! We've successfully eliminated the square root, and we're one step closer to finding the value(s) of 's'.

Step 3: Simplify and Rearrange into Quadratic Form

Now that we've squared both sides and expanded the equation, let's simplify and rearrange it into the standard quadratic form, which is ax² + bx + c = 0. This will make it easier for us to solve for 's'. Our current equation is:

s² - 8s + 16 = s + 2

To get it into the standard quadratic form, we need to move all terms to one side of the equation, leaving zero on the other side. Let's subtract 's' and 2 from both sides:

s² - 8s + 16 - s - 2 = 0

Now, combine like terms:

s² - 9s + 14 = 0

Fantastic! We now have a quadratic equation in standard form. This is a form we can easily work with using factoring, completing the square, or the quadratic formula. In the next step, we'll factor this quadratic equation to find the potential values for 's'.

Step 4: Factor the Quadratic Equation

We've got our quadratic equation in standard form: s² - 9s + 14 = 0. Now, let's factor it. Factoring involves finding two binomials that, when multiplied together, give us this quadratic equation. We're looking for two numbers that multiply to 14 (the constant term) and add up to -9 (the coefficient of the 's' term). Those numbers are -2 and -7 because (-2) * (-7) = 14 and (-2) + (-7) = -9. So, we can factor the equation as:

(s - 2)(s - 7) = 0

This factored form is incredibly useful because it tells us that the equation will be true if either (s - 2) = 0 or (s - 7) = 0. This is because anything multiplied by zero is zero. Now, we can easily solve for 's' by setting each factor equal to zero.

Step 5: Solve for s

From the factored equation (s - 2)(s - 7) = 0, we can find the potential solutions for 's'. We do this by setting each factor equal to zero:

• s - 2 = 0

  Adding 2 to both sides gives us:

  s = 2

• s - 7 = 0

  Adding 7 to both sides gives us:

  s = 7

So, we have two potential solutions: s = 2 and s = 7. But remember, because we squared both sides of the equation earlier, we might have introduced extraneous solutions. This means we need to check each of these potential solutions in the original equation to see if they actually work. This is a crucial step!

Step 6: Check for Extraneous Solutions

This is the most critical step! We need to check our potential solutions, s = 2 and s = 7, in the original equation: s = 4 + √(s + 2). This will tell us if both solutions are valid or if one (or both) are extraneous.

• Checking s = 2

  Substitute s = 2 into the original equation:

  2 = 4 + √(2 + 2)

  2 = 4 + √4

  2 = 4 + 2

  2 = 6

  This is not true. So, s = 2 is an extraneous solution.

• Checking s = 7

  Substitute s = 7 into the original equation:

  7 = 4 + √(7 + 2)

  7 = 4 + √9

  7 = 4 + 3

  7 = 7

  This is true. So, s = 7 is a valid solution.

We've done it! We checked both potential solutions and found that s = 2 doesn't work, but s = 7 does. This means s = 2 is an extraneous solution that we had to discard. This highlights why checking your answers is so important when dealing with square roots!

Final Answer

After going through all the steps – isolating the square root, squaring both sides, solving the quadratic equation, and most importantly, checking for extraneous solutions – we've arrived at our final answer. The only valid solution for the equation s = 4 + √(s + 2) is:

s = 7

That's it, guys! We successfully solved the equation. Remember, the key to solving equations with square roots is to isolate the square root, square both sides, and always, always, always check for extraneous solutions. This ensures you get the correct answer every time. Great job, and keep practicing those math skills!