Solve $\sqrt{3x+3}-1=x$: Step-by-Step Solution

Hey there, math enthusiasts! Let's dive into solving a radical equation. Radical equations, especially those involving square roots, can sometimes feel a bit tricky. But don't worry, we'll break it down step by step to make sure we get to the correct solution. Our mission today is to find the value(s) of x that satisfy the equation

$\sqrt{3x+3}-1=x$.

Before we jump into the solution, let's quickly review why these types of equations might need a little extra care. When we square both sides of an equation, we might introduce what are called extraneous solutions. These are values that pop up as potential answers but don't actually work when you plug them back into the original equation. So, it's super important to check our solutions at the end to make sure they're the real deal.

Step-by-Step Solution

1. Isolate the Radical

Our first move is to get the square root term all by itself on one side of the equation. To do this, we'll add 1 to both sides:

$\sqrt{3x+3} = x + 1$

2. Square Both Sides

Now that we have the square root isolated, we can get rid of it by squaring both sides of the equation. Remember, when we square the right side, we're squaring the entire expression (x + 1), not just x and 1 separately:

$(\sqrt{3x+3})^2 = (x + 1)^2$

This simplifies to:

$3x + 3 = x^2 + 2x + 1$

3. Rearrange into a Quadratic Equation

To solve for x, we need to get everything on one side of the equation, setting it equal to zero. This will give us a quadratic equation, which we can then solve using factoring, completing the square, or the quadratic formula. Let's subtract 3x and 3 from both sides:

$0 = x^2 + 2x + 1 - 3x - 3$

Simplifying, we get:

$0 = x^2 - x - 2$

4. Solve the Quadratic Equation

Now we have a quadratic equation in the form ax² + bx + c = 0. We can try to factor it first. We're looking for two numbers that multiply to -2 and add to -1. Those numbers are -2 and 1. So, we can factor the quadratic as follows:

$0 = (x - 2)(x + 1)$

Setting each factor equal to zero gives us potential solutions:

$x - 2 = 0$ or $x + 1 = 0$

Solving these, we find:

$x = 2$ or $x = -1$

5. Check for Extraneous Solutions

This is a crucial step! We need to plug our potential solutions back into the original equation to make sure they actually work.

Let's start with x = 2:

$\sqrt{3(2)+3} - 1 = 2$ $\sqrt{9} - 1 = 2$ $3 - 1 = 2$ $2 = 2$

So, x = 2 works!

Now let's check x = -1:

$\sqrt{3(-1)+3} - 1 = -1$ $\sqrt{0} - 1 = -1$ $0 - 1 = -1$ $-1 = -1$

So, x = -1 also works!

The Correct Answer

Both x = 2 and x = -1 satisfy the original equation. Therefore, the solution to the equation $\sqrt{3x+3} - 1 = x$ is:

A. -1 and 2

Why Checking for Extraneous Solutions is Important

Alright, guys, let's get real for a second about extraneous solutions. These sneaky little numbers can pop up when we're dealing with radical equations, and if we're not careful, they can lead us to the wrong answer. So, why do they happen?

The main reason we encounter extraneous solutions is because of the squaring process. When we square both sides of an equation, we're essentially saying that if a = b, then a² = b². While this is true, the reverse isn't always the case. If a² = b², it doesn't necessarily mean that a = b; it could also mean that a = -b.

Think about it this way: both 3² and (-3)² equal 9. So, if we start with x² = 9, the solutions could be x = 3 or x = -3. This is where the extra solution can creep in. When we're dealing with square roots, we're generally looking for the principal (positive) square root. But squaring both sides can introduce the negative possibility, which might not fit the original equation.

Let's illustrate this with a simple example. Suppose we have the equation:

$\sqrt{x} = -2$

Now, we know that the principal square root of a number can't be negative, so this equation actually has no real solutions. But let's see what happens if we square both sides:

$(\sqrt{x})^2 = (-2)^2$

This gives us:

$x = 4$

But if we plug x = 4 back into the original equation, we get:

$\sqrt{4} = -2$

$2 = -2$

This is clearly false! So, x = 4 is an extraneous solution.

In our original problem, checking for extraneous solutions was crucial because it confirmed that both x = 2 and x = -1 were valid. In other equations, you might find that one or more of your potential solutions don't work, and you'll need to discard them.

Here's a quick recap of why checking is so vital:

• Squaring can introduce extra solutions: As we've seen, squaring both sides can create solutions that don't actually satisfy the original equation.
• Square roots are generally positive: The principal square root is always non-negative, so negative results can indicate an extraneous solution.
• It ensures accuracy: Checking your solutions is the best way to make sure you've solved the equation correctly.

So, the next time you're tackling a radical equation, remember to always check your solutions. It's a small step that can save you from a big mistake!

Common Mistakes to Avoid

Alright, let's talk about some common pitfalls that students often encounter when solving equations like this. Knowing these mistakes can help you steer clear of them and ace your math problems!

1. Forgetting to Isolate the Radical First:

   This is a big one! Before you square both sides, you must isolate the radical term. If you don't, you'll end up with a much messier equation that's harder to solve. Remember, we want the square root term all by itself on one side of the equation. In our example, we added 1 to both sides to get $\sqrt{3x+3} = x + 1$. If we had squared both sides prematurely, we would have had to deal with $(\sqrt{3x+3} - 1)^2$, which is more complicated.

2. Squaring Terms Incorrectly:

   When squaring an expression with multiple terms, like (x + 1)², you need to use the distributive property (or the FOIL method) correctly. It's not just x² + 1²! The correct expansion is (x + 1)² = x² + 2x + 1. Missing the middle term (2x in this case) is a common mistake that can throw off your entire solution.

3. Skipping the Check for Extraneous Solutions:

   We've hammered this point home, but it's worth repeating: always check your solutions! Squaring both sides can introduce extraneous solutions, so plugging your potential answers back into the original equation is non-negotiable. If a solution doesn't work, toss it out!

4. Making Sign Errors:

   Sign errors are sneaky little devils that can trip you up at any stage of the problem. Be extra careful when moving terms across the equals sign and when applying the distributive property. A small sign mistake can lead to a completely wrong answer.

5. Factoring Quadratics Incorrectly:

   If your equation leads to a quadratic, make sure you factor it correctly. Double-check your factors to ensure they multiply to the correct constant term and add up to the correct coefficient of the x term. If factoring is giving you trouble, remember you can always use the quadratic formula.

6. Dividing by a Variable:

   This is a general algebra mistake, but it's worth mentioning. Never divide both sides of an equation by a variable unless you're sure that variable isn't zero. You might inadvertently eliminate a valid solution. For example, if you have x² = x, don't divide by x. Instead, rearrange the equation as x² - x = 0 and factor it as x(x - 1) = 0, which gives you the solutions x = 0 and x = 1.

7. Rushing Through the Problem:

   Math problems, especially those involving multiple steps, require focus and attention to detail. Rushing can lead to careless errors. Take your time, write neatly, and double-check each step. It's better to get the problem right than to finish quickly.

By being aware of these common mistakes, you can avoid them and boost your confidence in solving radical equations. Keep practicing, and you'll become a pro in no time!

Real-World Applications of Radical Equations

Okay, so we've conquered the steps to solve radical equations, and we know why it's crucial to check for those sneaky extraneous solutions. But you might be wondering, "Where does this stuff actually show up in the real world?" That's a fantastic question! Radical equations aren't just abstract math concepts; they have practical applications in various fields. Let's explore a few examples to see how these equations help us understand and solve real-world problems.

1. Physics: Calculating Projectile Motion

   Physics often involves understanding how objects move through space, and radical equations play a key role in analyzing projectile motion. For example, the time it takes for an object to fall a certain distance under the influence of gravity can be calculated using a radical equation. The formula looks something like this:

   $t = \sqrt{\frac{2h}{g}}$

   Where:

   • t is the time in seconds
   • h is the height in meters
   • g is the acceleration due to gravity (approximately 9.8 m/s²)

   So, if you want to know how long it will take a ball to fall from a 20-meter building, you'd plug the values into this equation and solve for t. The square root is essential here, making it a perfect example of a real-world application of radical equations.

2. Engineering: Designing Structures

   Engineers use radical equations to calculate various aspects of structural design, such as the strength and stability of bridges and buildings. For instance, the period of a pendulum (the time it takes for one complete swing) is given by the formula:

   $T = 2\pi\sqrt{\frac{L}{g}}$

   Where:

   • T is the period
   • L is the length of the pendulum
   • g is the acceleration due to gravity

   This equation helps engineers design structures that can withstand different forces and movements. Similarly, radical equations are used in calculating the resonant frequencies of structures, which is crucial for preventing collapses due to vibrations.

3. Astronomy: Understanding Orbital Motion

   In astronomy, radical equations help us understand the motion of celestial bodies. Kepler's Third Law of Planetary Motion, for example, relates the orbital period of a planet to the semi-major axis of its orbit. The equation involves a square root:

   $T^2 = \frac{4\pi^2a^3}{GM}$

   Where:

   • T is the orbital period
   • a is the semi-major axis
   • G is the gravitational constant
   • M is the mass of the central body (e.g., the Sun)

   This equation allows astronomers to calculate the orbital periods of planets and other celestial objects, helping us understand the dynamics of our solar system and the universe beyond.

4. Finance: Calculating Growth Rates

   Radical equations even show up in finance! Compound interest and growth rates often involve roots. For example, if you want to find the annual interest rate needed to double your investment in a certain number of years, you might use an equation that includes a root. These calculations are essential for financial planning and investment decisions.

5. Medicine: Dosage Calculations

   In medicine, certain dosage calculations can involve radical equations. For example, formulas used to determine the appropriate dosage of a medication based on a patient's weight or body surface area might include square roots. Accurate calculations are, of course, critical in healthcare to ensure patient safety.

These are just a few examples, guys, but they illustrate how radical equations are more than just abstract mathematical tools. They're essential for solving real-world problems in physics, engineering, astronomy, finance, medicine, and many other fields. So, the next time you're working on a radical equation, remember that you're learning a skill that has far-reaching applications!

Conclusion

So, there you have it! We've successfully solved the equation $\sqrt{3x+3} - 1 = x$, navigated the tricky waters of extraneous solutions, and highlighted some common mistakes to avoid. We've also seen how radical equations pop up in the real world, from physics to finance. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a master of radical equations in no time! And don't forget, always check your solutions!