Solving 7 = -2x^2 + 10x Complete Step-by-Step Guide

Jul 11, 2025 by ADMIN 52 views

Solving the Quadratic Equation 7 = -2x^2 + 10x: A Step-by-Step Guide

Solving quadratic equations is a fundamental skill in algebra, and understanding the step-by-step process is crucial for mastering this topic. In this article, we will walk through the complete steps for solving the quadratic equation $7 = -2x^2 + 10x$ . This comprehensive guide will cover factoring, completing the square, and expressing the trinomial as a binomial squared, ensuring you grasp each concept thoroughly. Whether you're a student tackling homework or someone looking to refresh their algebra skills, this article will provide a clear and detailed explanation.

Step 1: Rearrange the Equation and Factor

To effectively solve the quadratic equation $7 = -2x^2 + 10x$ , the first critical step involves rearranging the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$ . Rearranging the equation not only makes it easier to identify the coefficients but also sets the stage for applying various solution methods such as factoring, completing the square, or using the quadratic formula. In our case, we need to move all terms to one side of the equation, leaving zero on the other side. Starting with $7 = -2x^2 + 10x$ , we add $2x^2$ and subtract $10x$ from both sides to achieve this standard form. This transformation results in the equation $2x^2 - 10x + 7 = 0$ , which is now ready for further manipulation. This initial rearrangement is essential because it aligns the terms in a way that makes subsequent steps, such as factoring or completing the square, more straightforward and less prone to errors.

Now that the equation is in standard form, the next crucial step involves factoring out the coefficient of the $x^2$ term, if it's not equal to 1. Factoring simplifies the equation and prepares it for the method of completing the square. In our equation, $2x^2 - 10x + 7 = 0$ , the coefficient of the $x^2$ term is 2. We factor out 2 from the variable terms ( $2x^2$ and $-10x$ ), but we only focus on the terms that contain $x$ . This gives us $2(x^2 - 5x) + 7 = 0$ . Factoring out this coefficient is a pivotal step because it simplifies the process of completing the square, making the subsequent calculations more manageable. By factoring out the leading coefficient, we reduce the complexity of the equation, which is particularly helpful when the leading coefficient is not a perfect square. The resulting equation, $2(x^2 - 5x) + 7 = 0$ , is now in a more suitable form for the next steps in solving for $x$ .

Detailed Explanation of Rearranging and Factoring

The importance of rearranging the equation into the standard quadratic form, $ax^2 + bx + c = 0$ , cannot be overstated. This form provides a clear structure that allows for the easy identification of coefficients $a$ , $b$ , and $c$ , which are essential for applying various solution methods. For instance, the quadratic formula, $x = rac{-b ext{±} ext{√(}b^2 - 4ac)}{2a}$ , directly utilizes these coefficients. Similarly, completing the square and factoring methods are more efficiently applied when the equation is in standard form. Moreover, rearranging the equation ensures that all terms are accounted for and that no sign errors are made during the solution process. The transformation from $7 = -2x^2 + 10x$ to $2x^2 - 10x + 7 = 0$ is a foundational step that simplifies the overall problem. When factoring out the leading coefficient, in this case, 2, we create a quadratic expression inside the parentheses that is easier to complete the square. The expression $x^2 - 5x$ is much simpler to work with than $2x^2 - 10x$ . This simplification is key to avoiding fractions and maintaining accuracy throughout the solution. Furthermore, factoring out the coefficient prepares the equation for the next step, where we add and subtract a value inside the parentheses to create a perfect square trinomial. By focusing on the variable terms and leaving the constant term outside, we streamline the process and reduce the chances of errors. This careful and methodical approach ensures that the equation is in the best possible form for further manipulation, leading to a more accurate and efficient solution.

Step 2: Completing the Square

Completing the square is a powerful technique used to solve quadratic equations. After factoring out the leading coefficient, the next critical step is to complete the square within the parentheses. To do this, we need to add and subtract a specific value inside the parentheses that will create a perfect square trinomial. A perfect square trinomial is a quadratic expression that can be factored into the square of a binomial. The value we need to add and subtract is determined by taking half of the coefficient of the $x$ term, squaring it, and then adding and subtracting this value inside the parentheses. In our equation, $2(x^2 - 5x) + 7 = 0$ , the coefficient of the $x$ term is -5. Half of -5 is $- rac{5}{2}$ , and squaring this gives us $rac{25}{4}$ . Therefore, we add and subtract $rac{25}{4}$ inside the parentheses: $2(x^2 - 5x + rac{25}{4} - rac{25}{4}) + 7 = 0$ . This step is crucial because it transforms the quadratic expression into a form that can be easily factored as a binomial squared.

However, we must also account for the fact that we are adding and subtracting this value inside parentheses that are being multiplied by 2. So, when we add $rac{25}{4}$ inside the parentheses, we are actually adding $2 ext{\times} rac{25}{4} = rac{25}{2}$ to the left side of the equation. To maintain the balance of the equation, we must also subtract $rac{25}{2}$ on the left side. This means we effectively add $rac{25}{4}$ inside the parentheses and subtract $2 ext{\times} rac{25}{4}$ outside the parentheses. Thus, the equation becomes $2(x^2 - 5x + rac{25}{4}) + 7 - rac{25}{2} = 0$ . This adjustment ensures that we are not changing the overall value of the equation while creating the perfect square trinomial. Balancing the equation is a fundamental principle in algebra, and this step highlights the importance of careful bookkeeping when manipulating equations.

In-Depth Explanation of Completing the Square

Completing the square is a method that transforms a quadratic equation into a form where it can be easily solved. The core idea behind this technique is to create a perfect square trinomial, which is an expression of the form $(x + a)^2$ or $(x - a)^2$ . This transformation is achieved by adding and subtracting a specific value, derived from the coefficient of the $x$ term, inside the parentheses. The formula for finding this value is $( rac{b}{2})^2$ , where $b$ is the coefficient of the $x$ term. In our case, $b = -5$ , so we calculate $( rac{-5}{2})^2 = rac{25}{4}$ . Adding and subtracting this value allows us to rewrite the quadratic expression without changing its overall value while simultaneously creating a perfect square trinomial. The step of multiplying the added value by the leading coefficient and subtracting it outside the parentheses is crucial for maintaining the equation's balance. This ensures that we are only manipulating the form of the equation, not its solutions. For example, in the equation $2(x^2 - 5x + rac{25}{4} - rac{25}{4}) + 7 = 0$ , adding $rac{25}{4}$ inside the parentheses effectively adds $2 ext{\times} rac{25}{4} = rac{25}{2}$ to the left side. To counteract this, we subtract $rac{25}{2}$ outside the parentheses, resulting in $2(x^2 - 5x + rac{25}{4}) + 7 - rac{25}{2} = 0$ . This careful manipulation allows us to transform the equation into a solvable form while preserving its integrity. The resulting equation sets the stage for expressing the trinomial as a binomial squared and ultimately solving for $x$ .

Step 3: Write the Perfect Square Trinomial as a Binomial Squared

After completing the square, the next step is to express the perfect square trinomial as a binomial squared. This transformation simplifies the equation and brings us closer to solving for $x$ . In our equation, $2(x^2 - 5x + rac{25}{4}) + 7 - rac{25}{2} = 0$ , the expression inside the parentheses, $x^2 - 5x + rac{25}{4}$ , is a perfect square trinomial. This means it can be written in the form $(x - a)^2$ , where $a$ is half of the coefficient of the $x$ term. In this case, half of -5 is $- rac{5}{2}$ , so the perfect square trinomial can be written as $(x - rac{5}{2})^2$ . Therefore, the equation becomes $2(x - rac{5}{2})^2 + 7 - rac{25}{2} = 0$ . This step is crucial because it transforms a complex quadratic expression into a simpler, more manageable form. Writing the trinomial as a binomial squared allows us to isolate $x$ more easily in the subsequent steps.

Now, we simplify the constant terms outside the parentheses. We have $7 - rac{25}{2}$ . To combine these terms, we need a common denominator, which is 2. So, we rewrite 7 as $rac{14}{2}$ . Thus, the expression becomes $rac{14}{2} - rac{25}{2} = - rac{11}{2}$ . The equation now reads $2(x - rac{5}{2})^2 - rac{11}{2} = 0$ . This simplification makes the equation cleaner and easier to work with. By combining the constant terms, we prepare the equation for the final steps of solving for $x$ , which involve isolating the squared term and taking the square root. The transformation of the perfect square trinomial into a binomial squared, combined with the simplification of constant terms, is a critical milestone in solving the quadratic equation.

Detailed Explanation of Binomial Squared Transformation

The transformation of a perfect square trinomial into a binomial squared is a pivotal step in solving quadratic equations using the completing the square method. A perfect square trinomial, by definition, is a trinomial that can be factored into the form $(x + a)^2$ or $(x - a)^2$ . This form is incredibly useful because it simplifies the equation and allows us to isolate the variable $x$ . In our specific case, the trinomial $x^2 - 5x + rac{25}{4}$ fits this description perfectly. To rewrite it as a binomial squared, we recognize that the constant term, $rac{25}{4}$ , is the square of half the coefficient of the $x$ term, which is -5. Half of -5 is $- rac{5}{2}$ , and $(- rac{5}{2})^2 = rac{25}{4}$ . This confirms that the trinomial is indeed a perfect square. Therefore, we can rewrite $x^2 - 5x + rac{25}{4}$ as $(x - rac{5}{2})^2$ . This transformation significantly simplifies the equation, turning a complex expression into a concise and manageable form. Simplifying the constant terms outside the parentheses, such as combining $7$ and $- rac{25}{2}$ , is equally important. This involves finding a common denominator and performing the subtraction, which in our case results in $- rac{11}{2}$ . The equation $2(x - rac{5}{2})^2 - rac{11}{2} = 0$ is now in a streamlined format, ready for the subsequent steps of isolating the squared term and solving for $x$ . This process highlights the power of algebraic manipulation in simplifying complex problems and moving towards a solution.

Step 4: Solve for x

The final step in solving the quadratic equation is to isolate the squared term and then solve for $x$ . Starting from our simplified equation, $2(x - rac{5}{2})^2 - rac{11}{2} = 0$ , we first isolate the squared term by adding $rac{11}{2}$ to both sides of the equation. This gives us $2(x - rac{5}{2})^2 = rac{11}{2}$ . The next step is to divide both sides by 2 to further isolate the squared term. Dividing by 2, we get $(x - rac{5}{2})^2 = rac{11}{4}$ . Now, we have the squared term completely isolated, which sets the stage for taking the square root.

To remove the square, we take the square root of both sides of the equation. Remember that taking the square root introduces both positive and negative solutions. Therefore, we have $x - rac{5}{2} = ext{\pm} ext{\sqrt(} rac{11}{4})$ . We can simplify the square root by recognizing that $ ext√(}rac{11}{4}) = rac{ ext{√}11}{2}$. So, the equation becomes $x - rac{5}{2} = ext{\pm} rac{ ext{\sqrt}11}{2}$ . Finally, we solve for $x$ by adding $rac{5}{2}$ to both sides $x = rac{52} ext{±} rac{ ext{√}11}{2}$. This gives us two solutions for $x$ $x = rac{5 + ext{√11}{2}$ and $x = rac{5 - ext{\sqrt}11}{2}$ . These are the exact solutions to the quadratic equation. Solving for $x$ involves a series of algebraic manipulations that systematically isolate the variable, leading to the final answer.

Comprehensive Solution Explanation

The process of solving for $x$ after completing the square involves several key steps, each designed to isolate the variable and find its possible values. Starting with the equation $2(x - rac{5}{2})^2 - rac{11}{2} = 0$ , the first step is to isolate the term containing the squared expression. This is achieved by adding $rac{11}{2}$ to both sides of the equation, resulting in $2(x - rac{5}{2})^2 = rac{11}{2}$ . Next, we divide both sides by 2 to further isolate the squared term, which gives us $(x - rac{5}{2})^2 = rac{11}{4}$ . At this point, the squared expression is completely isolated, and we can proceed to take the square root of both sides. Taking the square root is a critical step because it introduces the possibility of both positive and negative solutions. This is represented by the ± symbol, indicating that there are two potential values for the expression inside the square root. Thus, we have $x - rac{5}{2} = ext{\pm} ext{\sqrt(} rac{11}{4})$ . Simplifying the square root, we get $x - rac{5}{2} = ext{\pm} rac{ ext{\sqrt}11}{2}$ . The final step is to isolate $x$ by adding $rac{5}{2}$ to both sides of the equation. This yields two solutions: $x = rac{5 + ext{\sqrt}11}{2}$ and $x = rac{5 - ext{\sqrt}11}{2}$ . These solutions represent the values of $x$ that satisfy the original quadratic equation. Each step in this process is carefully executed to ensure that the equation remains balanced and that the solutions are accurate. This systematic approach highlights the importance of algebraic manipulation in solving complex equations and arriving at the correct answers.

By following these steps, you can confidently solve the quadratic equation $7 = -2x^2 + 10x$ . Remember to rearrange the equation into standard form, factor out the leading coefficient, complete the square, express the trinomial as a binomial squared, and finally, solve for $x$ . This systematic approach will help you tackle various quadratic equations with ease.