Solve For \[$ Y \$\].$\[ \sqrt{y+2} + 3 = 8 \\]

Introduction

Solving equations with square roots can be a challenging task, but with the right approach, it can be broken down into manageable steps. In this article, we will focus on solving equations that involve square roots, specifically the equation $\sqrt{y+2} + 3 = 8$. We will use a step-by-step approach to solve for $y$ and provide a clear explanation of each step.

Understanding the Equation

The given equation is $\sqrt{y+2} + 3 = 8$. To solve for $y$, we need to isolate the square root term. The first step is to subtract 3 from both sides of the equation.

Subtracting 3 from Both Sides

$\sqrt{y+2} + 3 - 3 = 8 - 3$

$\sqrt{y+2} = 5$

Understanding the Square Root

The square root of a number is a value that, when multiplied by itself, gives the original number. In this case, the square root of $y+2$ is equal to 5. To find the value of $y$, we need to square both sides of the equation.

Squaring Both Sides

$(\sqrt{y+2})^2 = 5^2$

$y+2 = 25$

Solving for $y$

Now that we have the equation $y+2 = 25$, we can solve for $y$ by subtracting 2 from both sides.

Subtracting 2 from Both Sides

$y+2 - 2 = 25 - 2$

$y = 23$

Conclusion

In this article, we solved the equation $\sqrt{y+2} + 3 = 8$ by using a step-by-step approach. We started by subtracting 3 from both sides of the equation, then squared both sides to eliminate the square root. Finally, we solved for $y$ by subtracting 2 from both sides of the equation. The final answer is $y = 23$.

Tips and Tricks

• When solving equations with square roots, it's essential to isolate the square root term first.
• Squaring both sides of the equation can help eliminate the square root.
• Be careful when subtracting or adding numbers to both sides of the equation, as this can affect the final answer.

Real-World Applications

Solving equations with square roots has many real-world applications, such as:

• Calculating the area of a square or rectangle
• Finding the length of a side of a triangle
• Determining the volume of a cube or rectangular prism

Practice Problems

Try solving the following equations:

1. $\sqrt{x-4} + 2 = 9$
2. $\sqrt{y+1} - 3 = 4$
3. $\sqrt{z-2} + 1 = 6$

Solutions

1. $\sqrt{x-4} + 2 = 9$ $\sqrt{x-4} = 7$ $x-4 = 49$ $x = 53$
2. $\sqrt{y+1} - 3 = 4$ $\sqrt{y+1} = 7$ $y+1 = 49$ $y = 48$
3. $\sqrt{z-2} + 1 = 6$ $\sqrt{z-2} = 5$ $z-2 = 25$ $z = 27$

Conclusion

Introduction

In our previous article, we discussed how to solve equations with square roots using a step-by-step approach. In this article, we will answer some frequently asked questions (FAQs) related to solving equations with square roots.

Q: What is the first step in solving an equation with a square root?

A: The first step in solving an equation with a square root is to isolate the square root term. This means getting the square root term by itself on one side of the equation.

Q: How do I isolate the square root term?

A: To isolate the square root term, you can subtract or add numbers to both sides of the equation. For example, if the equation is $\sqrt{x} + 3 = 8$, you can subtract 3 from both sides to get $\sqrt{x} = 5$.

Q: What is the next step after isolating the square root term?

A: After isolating the square root term, the next step is to square both sides of the equation. This will eliminate the square root and allow you to solve for the variable.

Q: Why do I need to square both sides of the equation?

A: Squaring both sides of the equation is necessary to eliminate the square root. When you square both sides, you are essentially getting rid of the square root symbol and solving for the variable.

Q: What are some common mistakes to avoid when solving equations with square roots?

A: Some common mistakes to avoid when solving equations with square roots include:

• Not isolating the square root term first
• Not squaring both sides of the equation
• Not checking for extraneous solutions

Q: What is an extraneous solution?

A: An extraneous solution is a solution that is not valid or does not satisfy the original equation. When solving equations with square roots, it's essential to check for extraneous solutions to ensure that the solution is valid.

Q: How do I check for extraneous solutions?

A: To check for extraneous solutions, you can plug the solution back into the original equation and see if it's true. If the solution satisfies the original equation, then it's a valid solution.

Q: Can I use a calculator to solve equations with square roots?

A: Yes, you can use a calculator to solve equations with square roots. However, it's essential to understand the concept of square roots and how to solve equations with square roots manually.

Q: What are some real-world applications of solving equations with square roots?

A: Solving equations with square roots has many real-world applications, such as:

• Calculating the area of a square or rectangle
• Finding the length of a side of a triangle
• Determining the volume of a cube or rectangular prism

Q: Can I use solving equations with square roots to solve other types of equations?

A: Yes, solving equations with square roots can be used to solve other types of equations, such as quadratic equations and polynomial equations.

Conclusion

Solving equations with square roots requires a step-by-step approach and a clear understanding of the square root concept. By following the steps outlined in this article and avoiding common mistakes, you can solve equations with square roots and apply the concepts to real-world problems.

Practice Problems

Try solving the following equations:

1. $\sqrt{x} + 2 = 9$
2. $\sqrt{y} - 3 = 4$
3. $\sqrt{z} + 1 = 6$

Solutions

1. $\sqrt{x} + 2 = 9$ $\sqrt{x} = 7$ $x = 49$
2. $\sqrt{y} - 3 = 4$ $\sqrt{y} = 7$ $y = 49$
3. $\sqrt{z} + 1 = 6$ $\sqrt{z} = 5$ $z = 25$

Conclusion

Solving equations with square roots is a fundamental concept in algebra that has many real-world applications. By understanding the concept of square roots and following the steps outlined in this article, you can solve equations with square roots and apply the concepts to real-world problems.