Which Is A Solution To $(x-3)(x+9)=-27$?A. $x=-9$B. $ X = − 3 X=-3 X = − 3 [/tex]C. $x=0$D. $x=6$

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Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a specific quadratic equation, $(x-3)(x+9)=-27$, and explore the different methods and techniques used to find the solution.

Understanding the Equation

The given equation is a quadratic equation in the form of $(x-a)(x+b)=-c$. To solve this equation, we need to expand the left-hand side and set it equal to zero. This will allow us to use the quadratic formula or factorization to find the solutions.

Expanding the Equation

To expand the equation, we multiply the two binomials using the distributive property:

$$(x-3)(x+9) = x(x+9) - 3(x+9)$$

$$= x^2 + 9x - 3x - 27$$

$$= x^2 + 6x - 27$$

Setting the Equation Equal to Zero

Now that we have expanded the equation, we can set it equal to zero:

$$x^2 + 6x - 27 = 0$$

Using the Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It states that for an equation in the form of $ax^2 + bx + c = 0$, the solutions are given by:

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

In our case, $a = 1$, $b = 6$, and $c = -27$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-27)}}{2(1)}$$

$$= \frac{-6 \pm \sqrt{36 + 108}}{2}$$

$$= \frac{-6 \pm \sqrt{144}}{2}$$

$$= \frac{-6 \pm 12}{2}$$

Finding the Solutions

Now that we have the quadratic formula, we can find the solutions to the equation. We have two possible solutions:

$$x = \frac{-6 + 12}{2} = 3$$

$$x = \frac{-6 - 12}{2} = -9$$

Checking the Solutions

To verify that these solutions are correct, we can plug them back into the original equation:

$$(x-3)(x+9) = -27$$

For $x = 3$, we have:

$$(3-3)(3+9) = (0)(12) = 0 \neq -27$$

This means that $x = 3$ is not a solution to the equation.

For $x = -9$, we have:

$$( -9 -3)(-9+9) = (-12)(0) = 0 \neq -27$$

This means that $x = -9$ is not a solution to the equation.

Conclusion

In this article, we solved the quadratic equation $(x-3)(x+9)=-27$ using the quadratic formula. We found two possible solutions, $x = 3$ and $x = -9$, but upon checking, we found that neither of these solutions is correct. This highlights the importance of verifying solutions to ensure that they are accurate.

Final Answer

The correct solution to the equation $(x-3)(x+9)=-27$ is not among the options provided. However, if we re-examine the equation, we can see that it can be factored as:

$$(x-3)(x+9) = -27$$

$$(x-3)(x+9+27) = 0$$

$$(x-3)(x+36) = 0$$

This means that either $x-3 = 0$ or $x+36 = 0$. Solving for $x$, we get:

$$x = 3$$

$$x = -36$$

Therefore, the correct solution to the equation is $x = -36$.

Final Answer

The final answer is: $\boxed{-36}$

Introduction

Quadratic equations are a fundamental concept in mathematics, and solving them can be a challenging task for many students and professionals. In this article, we will address some of the most frequently asked questions about quadratic equations and provide detailed answers to help you better understand this topic.

Q1: What is a quadratic equation?

A1: A quadratic equation is a polynomial equation of degree two, which means that the highest power of the variable (usually x) is two. It is typically written in the form of ax^2 + bx + c = 0, where a, b, and c are constants.

Q2: How do I solve a quadratic equation?

A2: 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 your personal preference.

Q3: What is the quadratic formula?

A3: The quadratic formula is a mathematical formula that provides the solutions to a quadratic equation. It is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Q4: How do I use the quadratic formula?

A4: To use the quadratic formula, you need to identify the values of a, b, and c in the quadratic equation. Then, plug these values into the formula and simplify to find the solutions.

Q5: What is the difference between a quadratic equation and a linear equation?

A5: A quadratic equation is a polynomial equation of degree two, while a linear equation is a polynomial equation of degree one. In other words, a quadratic equation has a squared term, while a linear equation does not.

Q6: Can I solve a quadratic equation by factoring?

A6: Yes, you can solve a quadratic equation by factoring if it can be expressed as a product of two binomials. For example, the equation x^2 + 5x + 6 = 0 can be factored as (x + 3)(x + 2) = 0.

Q7: What is the significance of the discriminant in a quadratic equation?

A7: The discriminant is the expression under the square root in the quadratic formula, given by b^2 - 4ac. If the discriminant is positive, the equation has two distinct real solutions. If the discriminant is zero, the equation has one real solution. If the discriminant is negative, the equation has no real solutions.

Q8: Can I solve a quadratic equation using a calculator?

A8: Yes, you can solve a quadratic equation using a calculator. Most calculators have a built-in quadratic formula function that you can use to find the solutions.

Q9: What is the relationship between quadratic equations and conic sections?

A9: Quadratic equations are related to conic sections, which are geometric shapes that can be defined by a quadratic equation. For example, the equation x^2 + y^2 = 1 defines a circle, while the equation x^2 + y^2 = 4 defines a circle with a radius of 2.

Q10: Can I use quadratic equations to model real-world problems?

A10: Yes, quadratic equations can be used to model a wide range of real-world problems, including projectile motion, optimization problems, and electrical circuits.

Conclusion

Quadratic equations are a fundamental concept in mathematics, and solving them can be a challenging task. However, with practice and patience, you can master the techniques and methods used to solve quadratic equations. We hope that this Q&A article has provided you with a better understanding of quadratic equations and their applications.

Final Answer

The final answer is: Quadratic equations are a powerful tool for modeling and solving a wide range of real-world problems.