Solving $-x^2+2x+3=x^2-2x+3$ A Step-by-Step Guide

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In this article, we will delve into the process of finding the solution(s) for the quadratic equation βˆ’x2+2x+3=x2βˆ’2x+3-x^2 + 2x + 3 = x^2 - 2x + 3. Quadratic equations, which are polynomial equations of the second degree, play a crucial role in various fields, including mathematics, physics, engineering, and computer science. Understanding how to solve these equations is essential for tackling a wide range of problems. This article will provide a step-by-step guide to solving the given equation, ensuring a clear and comprehensive understanding of the methods involved. We'll explore the properties of quadratic equations, discuss the different approaches to solving them, and provide detailed explanations to make the process accessible to everyone, regardless of their mathematical background. The goal is to equip you with the knowledge and skills necessary to confidently solve similar equations in the future.

Understanding Quadratic Equations

Before diving into the solution, it’s important to understand the basics of quadratic equations. A quadratic equation is a polynomial equation of the second degree, meaning that the highest power of the variable is 2. The standard form of a quadratic equation is given by ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants, and a is not equal to 0. The solutions to a quadratic equation are also known as its roots or zeros, and they represent the values of x that satisfy the equation. These roots can be real or complex numbers, and a quadratic equation can have up to two distinct roots. The presence of the x2x^2 term is what distinguishes a quadratic equation from linear equations, which only have terms up to the first power of x. Recognizing this form is crucial for applying the appropriate methods to find the solutions. There are several methods for solving quadratic equations, including factoring, completing the square, and using the quadratic formula. Each method has its own advantages and is suitable for different types of quadratic equations. Understanding these methods and when to apply them is a key aspect of mastering quadratic equations. In the following sections, we will explore one such method to solve the given equation, breaking down each step for clarity and ease of understanding.

Simplifying the Equation

To begin, let's simplify the equation βˆ’x2+2x+3=x2βˆ’2x+3-x^2 + 2x + 3 = x^2 - 2x + 3. The first step is to consolidate all terms on one side of the equation to set it equal to zero. This is a standard practice in solving algebraic equations, as it allows us to work with a single expression and identify the coefficients more easily. By moving all terms to one side, we can rewrite the equation in the standard quadratic form, which makes it easier to apply various solution methods. In this case, we'll move the terms from the left side to the right side. We do this by adding x2x^2, subtracting 2x2x, and subtracting 3 from both sides of the equation. This gives us: 0=x2βˆ’2x+3+x2βˆ’2xβˆ’30 = x^2 - 2x + 3 + x^2 - 2x - 3. Now, we can combine like terms on the right side of the equation. We have x2+x2x^2 + x^2 which equals 2x22x^2, βˆ’2xβˆ’2x-2x - 2x which equals βˆ’4x-4x, and 3βˆ’33 - 3 which equals 0. So, the simplified equation becomes 2x2βˆ’4x=02x^2 - 4x = 0. This simplified form is easier to work with and allows us to identify potential solution methods more readily. The process of simplification is crucial in solving any equation, as it reduces complexity and makes the equation more manageable. In the next section, we will explore how to solve this simplified quadratic equation.

Solving 2x2βˆ’4x=02x^2 - 4x = 0

Now that we have the simplified equation 2x2βˆ’4x=02x^2 - 4x = 0, we can proceed to solve it. One effective method for solving this type of quadratic equation is by factoring. Factoring involves expressing the quadratic expression as a product of simpler expressions. In this case, we can factor out the greatest common factor (GCF) from both terms. The GCF of 2x22x^2 and βˆ’4x-4x is 2x2x. Factoring out 2x2x from the equation gives us 2x(xβˆ’2)=02x(x - 2) = 0. Now, we have a product of two factors equal to zero: 2x2x and (xβˆ’2)(x - 2). According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This gives us two separate equations to solve: 2x=02x = 0 and xβˆ’2=0x - 2 = 0. Solving the first equation, 2x=02x = 0, we divide both sides by 2 to get x=0x = 0. Solving the second equation, xβˆ’2=0x - 2 = 0, we add 2 to both sides to get x=2x = 2. Therefore, the solutions to the quadratic equation 2x2βˆ’4x=02x^2 - 4x = 0 are x=0x = 0 and x=2x = 2. Factoring is a powerful technique for solving quadratic equations, especially when the equation can be easily factored. It provides a direct and efficient way to find the roots of the equation. In the next section, we will summarize the solutions and discuss their significance.

Solutions and Conclusion

In conclusion, by simplifying the original equation βˆ’x2+2x+3=x2βˆ’2x+3-x^2 + 2x + 3 = x^2 - 2x + 3 and then solving the resulting quadratic equation 2x2βˆ’4x=02x^2 - 4x = 0, we found two distinct solutions: x=0x = 0 and x=2x = 2. These solutions represent the values of x that satisfy the original equation. To verify these solutions, we can substitute them back into the original equation and check if the equation holds true. For x=0x = 0, we have βˆ’02+2(0)+3=02βˆ’2(0)+3-0^2 + 2(0) + 3 = 0^2 - 2(0) + 3, which simplifies to 3=33 = 3, confirming that x=0x = 0 is a solution. For x=2x = 2, we have βˆ’(2)2+2(2)+3=(2)2βˆ’2(2)+3-(2)^2 + 2(2) + 3 = (2)^2 - 2(2) + 3, which simplifies to βˆ’4+4+3=4βˆ’4+3-4 + 4 + 3 = 4 - 4 + 3, and further to 3=33 = 3, confirming that x=2x = 2 is also a solution. Thus, the solutions x=0x = 0 and x=2x = 2 are indeed the roots of the equation. Understanding how to solve quadratic equations is a fundamental skill in algebra and has wide-ranging applications in various fields. The method of factoring, as demonstrated in this article, is a powerful tool for solving quadratic equations that can be easily factored. By mastering this technique, you can confidently tackle a variety of mathematical problems and real-world applications involving quadratic equations.