Select The Correct Answer From The Drop-down Menu.Find The Solution Set.The Solution Set For $5v^2 - 125 = 0$ Is $\square$.

Mar 11, 2025 by ADMIN 128 views

**Solving Quadratic Equations: Finding the Solution Set**

===========================================================

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we will focus on solving quadratic equations of the form $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants. We will use the quadratic formula to find the solution set for a given equation.

Understanding Quadratic Equations

A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants. The variable in the equation can be any letter, but in this case, we will use $v$ .

The Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

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

This formula will give us two solutions for the equation, which are the values of $x$ that satisfy the equation.

Solving the Given Equation

The given equation is $5v^2 - 125 = 0$ . To solve this equation, we need to isolate the variable $v$ . We can start by adding 125 to both sides of the equation:

5v^2 = 125

Next, we can divide both sides of the equation by 5:

v^2 = 25

Now, we can take the square root of both sides of the equation:

v = \pm \sqrt{25}

Simplifying the square root, we get:

v = \pm 5

Therefore, the solution set for the equation $5v^2 - 125 = 0$ is $\boxed{\{-5, 5\}}$ .

Interpretation of the Solution Set

The solution set for the equation $5v^2 - 125 = 0$ is $\{-5, 5\}$ . This means that the values of $v$ that satisfy the equation are $-5$ and $5$ . In other words, when we substitute $-5$ or $5$ for $v$ in the original equation, the equation will be true.

Conclusion

Solving quadratic equations is an essential skill for students to master. In this article, we used the quadratic formula to find the solution set for the equation $5v^2 - 125 = 0$ . We started by adding 125 to both sides of the equation, then divided both sides by 5, and finally took the square root of both sides. The solution set for the equation is $\{-5, 5\}$ , which means that the values of $v$ that satisfy the equation are $-5$ and $5$ .

Example Problems

Problem 1

Solve the equation $2x^2 + 5x - 3 = 0$ .

Solution

To solve this equation, we can use the quadratic formula:

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

In this case, $a = 2$ , $b = 5$ , and $c = -3$ . Plugging these values into the formula, we get:

x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}

Simplifying the expression under the square root, we get:

x = \frac{-5 \pm \sqrt{25 + 24}}{4}

x = \frac{-5 \pm \sqrt{49}}{4}

x = \frac{-5 \pm 7}{4}

Therefore, the solutions to the equation are $x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}$ and $x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3$ .

Problem 2

Solve the equation $x^2 - 4x + 4 = 0$ .

Solution

To solve this equation, we can use the quadratic formula:

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

In this case, $a = 1$ , $b = -4$ , and $c = 4$ . Plugging these values into the formula, we get:

x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)}

Simplifying the expression under the square root, we get:

x = \frac{4 \pm \sqrt{16 - 16}}{2}

x = \frac{4 \pm \sqrt{0}}{2}

x = \frac{4}{2}

Therefore, the solution to the equation is $x = \frac{4}{2} = 2$ .

Tips and Tricks

Tip 1

When solving quadratic equations, make sure to simplify the expression under the square root before plugging it into the quadratic formula.

Tip 2

When using the quadratic formula, make sure to plug in the correct values for $a$ , $b$ , and $c$ .

Tip 3

When solving quadratic equations, make sure to check your solutions by plugging them back into the original equation.

Conclusion

===========================================================

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we will answer some of the most frequently asked questions about quadratic equations.

Q: What is a quadratic equation?

A: A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is two. The general form of a quadratic equation is $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are constants.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the quadratic formula:

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

This formula will give you two solutions for the equation, which are the values of $x$ that satisfy the equation.

Q: What is the quadratic formula?

A: The quadratic formula is a powerful tool for solving quadratic equations. It is given by:

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

This formula will give you two solutions for the equation, which are the values of $x$ that satisfy the 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. Then, simplify the expression under the square root and solve for $x$ .

Q: What is the difference between the quadratic formula and factoring?

A: The quadratic formula and factoring are two different methods for solving quadratic equations. The quadratic formula is a general method that works for all quadratic equations, while factoring is a specific method that only works for certain types of quadratic equations.

Q: How do I know which method to use?

A: To determine which method to use, you need to look at the equation and see if it can be factored. If it can be factored, then you can use the factoring method. If it cannot be factored, then you need to use the quadratic formula.

Q: What is the solution set for a quadratic equation?

A: The solution set for a quadratic equation is the set of all values of $x$ that satisfy the equation. This can be a single value, two values, or no values at all.

Q: How do I find the solution set for a quadratic equation?

A: To find the solution set for a quadratic equation, you need to use the quadratic formula and solve for $x$ . Then, simplify the expression and find the values of $x$ that satisfy the equation.

Q: What is the discriminant in the quadratic formula?

A: The discriminant in the quadratic formula is the expression under the square root, which is $b^2 - 4ac$ . This expression determines the nature of the solutions to the equation.

Q: How do I determine the nature of the solutions to a quadratic equation?

A: To determine the nature of the solutions to a quadratic equation, you need to look at the discriminant. If the discriminant is positive, then the equation has two distinct real solutions. If the discriminant is zero, then the equation has one real solution. If the discriminant is negative, then the equation has no real solutions.

Q: What is the difference between a quadratic equation and a linear equation?

A: A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. This means that a quadratic equation has a highest power of two, while a linear equation has a highest power of one.

Q: How do I solve a linear equation?

A: To solve a linear equation, you can use the following steps:

Add or subtract the same value to both sides of the equation to isolate the variable.
Multiply or divide both sides of the equation by the same value to solve for the variable.

Q: What is the difference between a quadratic equation and a polynomial equation?

A: A quadratic equation is a polynomial equation of degree two, while a polynomial equation is a general term that refers to any equation with a variable raised to a power. This means that a quadratic equation is a specific type of polynomial equation.

Q: How do I solve a polynomial equation?

A: To solve a polynomial equation, you need to use the following steps:

Factor the equation, if possible.
Use the quadratic formula, if the equation is quadratic.
Use other methods, such as synthetic division or the rational root theorem, if the equation is not quadratic.

Conclusion

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students to master. In this article, we answered some of the most frequently asked questions about quadratic equations, including how to solve them, what the quadratic formula is, and how to determine the nature of the solutions to a quadratic equation. We also discussed the difference between quadratic equations and linear equations, and how to solve polynomial equations.