Solve The Equation \[$(-3x - 15)(x + 7) = 0\$\].A. 6 And 7 B. -5 And -7 C. -5 And 6 D. 3 And 9

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Introduction

In mathematics, equations are a fundamental concept that help us understand and describe various relationships between variables. Solving equations is a crucial skill that enables us to find the values of unknown variables. In this article, we will focus on solving a quadratic equation of the form (βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0}. We will break down the solution process into manageable steps and provide a clear explanation of each step.

Understanding the Equation

The given equation is a quadratic equation in the form of a product of two binomials. To solve this equation, we need to find the values of x that make the equation true. In other words, we need to find the values of x that make the product of the two binomials equal to zero.

Step 1: Apply the Zero Product Property

The zero product property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero. In this case, we can apply the zero product property to the given equation:

(βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0}

This means that either βˆ’3xβˆ’15=0{-3x - 15 = 0} or x+7=0{x + 7 = 0}.

Step 2: Solve the First Equation

Let's start by solving the first equation:

βˆ’3xβˆ’15=0{-3x - 15 = 0}

To solve for x, we can add 15 to both sides of the equation:

βˆ’3x=15{-3x = 15}

Next, we can divide both sides of the equation by -3:

x=βˆ’5{x = -5}

Step 3: Solve the Second Equation

Now, let's solve the second equation:

x+7=0{x + 7 = 0}

To solve for x, we can subtract 7 from both sides of the equation:

x=βˆ’7{x = -7}

Conclusion

In conclusion, we have solved the given quadratic equation (βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0} using the zero product property. We found that the solutions to the equation are x = -5 and x = -7.

Answer

The correct answer is:

B. -5 and -7

Why is this the correct answer?

This is the correct answer because we have shown that x = -5 and x = -7 are the solutions to the given equation. These values of x make the equation true, and therefore, they are the correct solutions.

Tips and Tricks

  • When solving quadratic equations, it's essential to apply the zero product property to find the values of x that make the equation true.
  • Make sure to simplify the equation and isolate the variable x before solving for its value.
  • Check your solutions by plugging them back into the original equation to ensure that they are true.

Common Mistakes

  • Failing to apply the zero product property when solving quadratic equations.
  • Not simplifying the equation and isolating the variable x before solving for its value.
  • Not checking the solutions by plugging them back into the original equation.

Real-World Applications

Solving quadratic equations has numerous real-world applications in various fields, including physics, engineering, and economics. For example, in physics, quadratic equations are used to model the motion of objects under the influence of gravity. In engineering, quadratic equations are used to design and optimize systems, such as bridges and buildings. In economics, quadratic equations are used to model the behavior of markets and predict future trends.

Conclusion

Introduction

In our previous article, we solved the quadratic equation (βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0} using the zero product property. In this article, we will provide a Q&A guide to help you understand the solution process and address any questions you may have.

Q: What is the zero product property?

A: The zero product property states that if the product of two or more factors is equal to zero, then at least one of the factors must be equal to zero.

Q: How do I apply the zero product property to solve a quadratic equation?

A: To apply the zero product property, you need to set each factor equal to zero and solve for x. In the case of the equation (βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0}, we set each factor equal to zero and solve for x:

βˆ’3xβˆ’15=0{-3x - 15 = 0}

x+7=0{x + 7 = 0}

Q: What if I have a quadratic equation with a coefficient of 1?

A: If you have a quadratic equation with a coefficient of 1, you can simply factor the equation and set each factor equal to zero. For example, consider the equation x2+5x+6=0{x^2 + 5x + 6 = 0}. You can factor the equation as:

(x+3)(x+2)=0{(x + 3)(x + 2) = 0}

Then, you can set each factor equal to zero and solve for x:

x+3=0{x + 3 = 0}

x+2=0{x + 2 = 0}

Q: What if I have a quadratic equation with a coefficient of -1?

A: If you have a quadratic equation with a coefficient of -1, you can simply factor the equation and set each factor equal to zero. For example, consider the equation βˆ’x2βˆ’4xβˆ’4=0{-x^2 - 4x - 4 = 0}. You can factor the equation as:

βˆ’(x+2)(x+2)=0{-(x + 2)(x + 2) = 0}

Then, you can set each factor equal to zero and solve for x:

βˆ’(x+2)=0{-(x + 2) = 0}

Q: How do I check my solutions?

A: To check your solutions, you need to plug them back into the original equation and verify that they are true. For example, consider the equation (βˆ’3xβˆ’15)(x+7)=0{(-3x - 15)(x + 7) = 0}. If you found that x = -5 is a solution, you can plug x = -5 back into the equation and verify that it is true:

(βˆ’3(βˆ’5)βˆ’15)(βˆ’5+7)=0{(-3(-5) - 15)(-5 + 7) = 0}

15(βˆ’2)=0{15(-2) = 0}

30=0{30 = 0}

This is not true, so x = -5 is not a solution. However, if you found that x = -7 is a solution, you can plug x = -7 back into the equation and verify that it is true:

(βˆ’3(βˆ’7)βˆ’15)(βˆ’7+7)=0{(-3(-7) - 15)(-7 + 7) = 0}

21(βˆ’0)=0{21(-0) = 0}

0=0{0 = 0}

This is true, so x = -7 is a solution.

Q: What if I have a quadratic equation with no real solutions?

A: If you have a quadratic equation with no real solutions, it means that the equation has complex solutions. Complex solutions are solutions that involve imaginary numbers, such as i. For example, consider the equation x2+1=0{x^2 + 1 = 0}. You can factor the equation as:

(x+i)(xβˆ’i)=0{(x + i)(x - i) = 0}

Then, you can set each factor equal to zero and solve for x:

x+i=0{x + i = 0}

xβˆ’i=0{x - i = 0}

The solutions to this equation are x = -i and x = i.

Conclusion

In conclusion, solving quadratic equations is a crucial skill that has numerous real-world applications. By applying the zero product property and simplifying the equation, we can find the values of x that make the equation true. Remember to check your solutions by plugging them back into the original equation to ensure that they are true. With practice and patience, you can become proficient in solving quadratic equations and apply them to real-world problems.