What Is The Solution To The Equation?${ \sqrt{x-3} + 1 = 6 }$A. 64 B. 46 C. 28 D. 22

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Introduction

Solving equations is a fundamental concept in mathematics, and it requires a step-by-step approach to arrive at the correct solution. In this article, we will focus on solving a specific equation involving a square root. The equation is xβˆ’3+1=6\sqrt{x-3} + 1 = 6. Our goal is to isolate the variable xx and find its value.

Understanding the Equation

The given equation involves a square root, which can be solved by isolating the square root term and then squaring both sides of the equation. The equation is xβˆ’3+1=6\sqrt{x-3} + 1 = 6. To start solving this equation, we need to isolate the square root term.

Isolating the Square Root Term

To isolate the square root term, we need to subtract 1 from both sides of the equation. This will give us xβˆ’3=5\sqrt{x-3} = 5. Now, we have isolated the square root term, and we can proceed to the next step.

Squaring Both Sides of the Equation

Squaring both sides of the equation is a crucial step in solving equations involving square roots. When we square both sides of the equation xβˆ’3=5\sqrt{x-3} = 5, we get xβˆ’3=25x-3 = 25. This is because the square of a square root is equal to the number inside the square root.

Solving for x

Now that we have xβˆ’3=25x-3 = 25, we can solve for xx by adding 3 to both sides of the equation. This gives us x=28x = 28. Therefore, the solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 is x=28x = 28.

Conclusion

In this article, we solved a specific equation involving a square root. We started by isolating the square root term and then squared both sides of the equation to arrive at the solution. The solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 is x=28x = 28. This demonstrates the importance of following a step-by-step approach when solving equations.

Frequently Asked Questions

  • What is the solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6?
  • How do you isolate the square root term in an equation?
  • What is the next step after isolating the square root term in an equation?

Answer Key

  • The solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 is x=28x = 28.
  • To isolate the square root term, you need to subtract the constant term from both sides of the equation.
  • The next step after isolating the square root term is to square both sides of the equation.

Step-by-Step Solution

  1. Isolate the square root term by subtracting 1 from both sides of the equation.
  2. Square both sides of the equation to eliminate the square root term.
  3. Solve for xx by adding 3 to both sides of the equation.

Tips and Tricks

  • When solving equations involving square roots, make sure to isolate the square root term first.
  • Squaring both sides of the equation is a crucial step in solving equations involving square roots.
  • Always check your solution by plugging it back into the original equation.

Real-World Applications

Solving equations involving square roots has numerous real-world applications. For example, in physics, the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 can be used to model the motion of an object. In finance, the equation can be used to calculate the value of an investment. In engineering, the equation can be used to design a system that involves square roots.

Conclusion

In conclusion, solving equations involving square roots requires a step-by-step approach. By isolating the square root term and then squaring both sides of the equation, we can arrive at the solution. The solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 is x=28x = 28. This demonstrates the importance of following a step-by-step approach when solving equations.

Final Answer

The final answer is 28\boxed{28}.

Introduction

Solving equations involving square roots can be a challenging task, but with the right approach, it can be made easier. In this article, we will answer some of the most frequently asked questions related to solving equations involving square roots.

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

A: The first step in solving an equation involving a square root is to isolate the square root term. This can be done by subtracting the constant term from both sides of the equation.

Q: How do I isolate the square root term in an equation?

A: To isolate the square root term, you need to subtract the constant term from both sides of the equation. For example, in the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6, you would subtract 1 from both sides to get xβˆ’3=5\sqrt{x-3} = 5.

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

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

Q: How do I square both sides of an equation?

A: To square both sides of an equation, you simply multiply both sides by themselves. For example, in the equation xβˆ’3=5\sqrt{x-3} = 5, you would square both sides to get xβˆ’3=25x-3 = 25.

Q: How do I solve for the variable in an equation involving a square root?

A: To solve for the variable in an equation involving a square root, you need to isolate the variable on one side of the equation. This can be done by adding or subtracting the constant term from both sides of the equation.

Q: What is the final step in solving an equation involving a square root?

A: The final step in solving an equation involving a square root is to check your solution by plugging it back into the original equation. This will ensure that your solution is correct.

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

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

  • Not isolating the square root term
  • Not squaring both sides of the equation
  • Not checking the solution by plugging it back into the original equation

Q: How do I check my solution by plugging it back into the original equation?

A: To check your solution by plugging it back into the original equation, you simply substitute the solution into the original equation and see if it is true. For example, if the solution to the equation xβˆ’3+1=6\sqrt{x-3} + 1 = 6 is x=28x = 28, you would plug x=28x = 28 into the original equation to get 28βˆ’3+1=6\sqrt{28-3} + 1 = 6. If this is true, then your solution is correct.

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

A: Solving equations involving square roots has numerous real-world applications, including:

  • Modeling the motion of an object in physics
  • Calculating the value of an investment in finance
  • Designing a system that involves square roots in engineering

Conclusion

In conclusion, solving equations involving square roots requires a step-by-step approach. By isolating the square root term, squaring both sides of the equation, and checking the solution, you can arrive at the correct solution. We hope that this article has answered some of the most frequently asked questions related to solving equations involving square roots.

Final Answer

The final answer is 28\boxed{28}.