Using the Square Root Property

Oct 13, 2025 by wbrum.org 62 views

Introduction

Quadratic equations are a fundamental concept in algebra, and solving them algebraically is a crucial skill for students to master. In this article, we will focus on solving the quadratic equation $x^2 = 36$ algebraically, and we will explore the different methods and techniques used to solve this type of equation.

What is a Quadratic Equation?

A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable (in this case, $x$ ) is two. The general form of a quadratic equation is:

ax^2 + bx + c = 0

where $a$ , $b$ , and $c$ are constants, and $x$ is the variable.

Solving the Equation $x^2 = 36$

To solve the equation $x^2 = 36$ , we need to find the values of $x$ that satisfy this equation. We can start by taking the square root of both sides of the equation:

x^2 = 36

x = \pm \sqrt{36}

Using the Square Root Property

The square root property states that if $x^2 = a$ , then $x = \pm \sqrt{a}$ . In this case, we have:

x = \pm \sqrt{36}

Since $\sqrt{36} = 6$ , we can simplify the equation to:

x = \pm 6

Checking the Solutions

To check our solutions, we can plug them back into the original equation:

x^2 = 36

If we substitute $x = 6$ into the equation, we get:

(6)^2 = 36

36 = 36

This is true, so $x = 6$ is a solution to the equation.

If we substitute $x = -6$ into the equation, we get:

(-6)^2 = 36

36 = 36

This is also true, so $x = -6$ is also a solution to the equation.

Conclusion

In this article, we have solved the quadratic equation $x^2 = 36$ algebraically using the square root property. We have found that the solutions to this equation are $x = 6$ and $x = -6$ . These solutions satisfy the original equation, and they can be verified by plugging them back into the equation.

Final Answer

The final answer is: B

Discussion

This problem is a great example of how to use the square root property to solve quadratic equations. The square root property is a powerful tool that can be used to solve equations of the form $x^2 = a$ . By using this property, we can find the solutions to the equation $x^2 = 36$ and verify that they satisfy the original equation.

Additional Resources

For more information on solving quadratic equations, check out the following resources:

Khan Academy: Solving Quadratic Equations
Mathway: Solving Quadratic Equations
Wolfram Alpha: Solving Quadratic Equations

Introduction

Quadratic equations are a fundamental concept in algebra, and solving them can be a challenging task for many students. In this article, we will answer some of the most frequently asked questions about quadratic equations, including how to solve them, what the solutions mean, and how to apply them in real-world situations.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable (in this case, $x$ ) is two. The general form of a quadratic equation is:

ax^2 + bx + c = 0

where $a$ , $b$ , and $c$ are constants, and $x$ is the variable.

Q: How do I solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including factoring, using the quadratic formula, and completing the square. The method you choose will depend on the specific equation and the type of solutions you are looking for.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that can be used to solve quadratic equations. It is given by:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

Q: How do I use the quadratic formula?

A: To use the quadratic formula, you need to plug in the values of $a$ , $b$ , and $c$ into the formula. You will then get two solutions for $x$ , which are given by:

x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}

and

x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

Q: What do the solutions to a quadratic equation mean?

A: The solutions to a quadratic equation represent the values of $x$ that satisfy the equation. In other words, they are the values of $x$ that make the equation true.

Q: How do I apply quadratic equations in real-world situations?

A: Quadratic equations have many real-world applications, including physics, engineering, and economics. For example, quadratic equations can be used to model the motion of objects under the influence of gravity, to calculate the trajectory of a projectile, or to determine the maximum or minimum value of a function.

Q: What are some common mistakes to avoid when solving quadratic equations?

A: Some common mistakes to avoid when solving quadratic equations include:

Not checking the solutions to see if they are valid
Not using the correct method to solve the equation
Not simplifying the solutions to their simplest form
Not checking the solutions to see if they satisfy the original equation

Q: How do I check if a solution is valid?

A: To check if a solution is valid, you need to plug it back into the original equation and see if it satisfies the equation. If it does, then the solution is valid. If it doesn't, then the solution is not valid.

Conclusion

In this article, we have answered some of the most frequently asked questions about quadratic equations, including how to solve them, what the solutions mean, and how to apply them in real-world situations. We hope that this article has been helpful in clarifying any confusion you may have had about quadratic equations.

Final Answer

The final answer is: There is no final answer, as this is a Q&A article.

Discussion

This article is a great resource for anyone who wants to learn more about quadratic equations. We hope that you have found the information in this article to be helpful and informative.

Additional Resources

For more information on quadratic equations, check out the following resources:

Khan Academy: Quadratic Equations
Mathway: Quadratic Equations
Wolfram Alpha: Quadratic Equations