Solving $2x(8-7x)=-11$ A Detailed Guide To Quadratic Equations

Introduction

In this article, we will delve into the process of solving a quadratic equation presented in the form $2x(8-7x) = -11$. Quadratic equations, which are polynomial equations of the second degree, frequently appear in various fields of mathematics, physics, engineering, and other scientific disciplines. The general form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants, and $x$ represents the variable we aim to find. Solving such equations involves determining the values of $x$ that satisfy the equation. In many real-world applications, finding accurate solutions is crucial for precise calculations and predictions. This article will not only guide you through the steps to solve the given equation but also emphasize the importance of understanding the underlying mathematical concepts. We will cover the process of expanding and rearranging the equation into the standard quadratic form, identifying the coefficients, and then applying the quadratic formula to find the solutions. Additionally, we will round the solutions to two decimal places as requested, ensuring that the final answers are both accurate and practical for real-world use. This detailed explanation will help you grasp the fundamental principles of solving quadratic equations and appreciate their wide-ranging applications.

Expanding and Rearranging the Equation

To begin, our initial equation is $2x(8-7x) = -11$. The first step in solving this equation is to expand the left side by distributing $2x$ across the terms inside the parenthesis. This process involves multiplying $2x$ by both $8$ and $-7x$. When we multiply $2x$ by $8$, we obtain $16x$. Similarly, when we multiply $2x$ by $-7x$, we get $-14x^2$. Thus, the expanded form of the left side of the equation is $16x - 14x^2$. Now, our equation looks like $16x - 14x^2 = -11$. The next step is to rearrange the equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. To achieve this, we need to move all the terms to one side of the equation, leaving zero on the other side. We can do this by adding $11$ to both sides of the equation. This gives us $-14x^2 + 16x + 11 = 0$. It is conventional to have the coefficient of $x^2$ as a positive number. To accomplish this, we multiply the entire equation by $-1$. This changes the signs of all the terms, resulting in the equation $14x^2 - 16x - 11 = 0$. Now, the equation is in the standard quadratic form, which makes it easier to identify the coefficients $a$, $b$, and $c$ that we will use in the quadratic formula. This rearrangement is crucial because it sets the stage for applying the quadratic formula and finding the solutions for $x$. Understanding this process ensures that you can tackle any quadratic equation effectively.

Identifying Coefficients and Applying the Quadratic Formula

After rearranging the equation into the standard quadratic form $14x^2 - 16x - 11 = 0$, the next critical step is to identify the coefficients $a$, $b$, and $c$. In this equation, the coefficient of $x^2$ is $a$, the coefficient of $x$ is $b$, and the constant term is $c$. By comparing the equation with the standard form $ax^2 + bx + c = 0$, we can see that $a = 14$, $b = -16$, and $c = -11$. These coefficients are essential because they are used in the quadratic formula, which is a general method for solving quadratic equations. The quadratic formula is given by:

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

This formula provides the two possible solutions for $x$ based on the values of $a$, $b$, and $c$. The $\pm$ symbol indicates that there are two solutions: one where we add the square root term and one where we subtract it. Now, we substitute the identified coefficients into the formula. Plugging in $a = 14$, $b = -16$, and $c = -11$, we get:

$$x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(14)(-11)}}{2(14)}$$

This substitution is a crucial step, and ensuring its accuracy is vital for obtaining the correct solutions. The next step involves simplifying the expression under the square root and the rest of the formula. Understanding how to correctly substitute and simplify the equation is key to successfully solving quadratic equations. This process demonstrates the power and versatility of the quadratic formula in finding solutions for a wide range of quadratic equations.

Simplifying the Quadratic Formula

Having substituted the coefficients into the quadratic formula, the next crucial step is simplifying the expression to find the values of $x$. The formula now looks like this:

$$x = \frac{16 \pm \sqrt{(-16)^2 - 4(14)(-11)}}{2(14)}$$

First, let's simplify the expression under the square root. We have $(-16)^2$, which is $(-16) \times (-16) = 256$. Next, we calculate $-4(14)(-11)$. Multiplying these numbers gives us $-4 \times 14 \times -11 = 616$. So, the expression under the square root becomes $256 + 616$, which equals $872$. Now our formula looks like this:

$$x = \frac{16 \pm \sqrt{872}}{28}$$

Next, we need to find the square root of $872$. The square root of $872$ is approximately $29.53$. Therefore, our equation becomes:

$$x = \frac{16 \pm 29.53}{28}$$

Now we have two possible solutions for $x$, one where we add $29.53$ to $16$ and another where we subtract $29.53$ from $16$. This simplification process is critical because it breaks down the complex formula into manageable parts, making it easier to arrive at the final solutions. The ability to accurately simplify such expressions is a fundamental skill in solving quadratic equations and other mathematical problems. By understanding each step in this simplification, you can confidently tackle more complex equations.

Calculating the Two Solutions and Rounding to 2 Decimal Places

After simplifying the quadratic formula, we arrived at the expression:

$$x = \frac{16 \pm 29.53}{28}$$

This expression gives us two potential solutions for $x$. Let's calculate them one by one. First, we'll consider the case where we add $29.53$ to $16$:

$$x_1 = \frac{16 + 29.53}{28}$$

Adding the numbers in the numerator gives us $16 + 29.53 = 45.53$. Now, we divide this sum by $28$:

$$x_1 = \frac{45.53}{28} \approx 1.626$$

Rounding this to two decimal places, we get $x_1 \approx 1.63$. Now, let's calculate the second solution, where we subtract $29.53$ from $16$:

$$x_2 = \frac{16 - 29.53}{28}$$

Subtracting the numbers in the numerator gives us $16 - 29.53 = -13.53$. Now, we divide this difference by $28$:

$$x_2 = \frac{-13.53}{28} \approx -0.483$$

Rounding this to two decimal places, we get $x_2 \approx -0.48$. Therefore, the two solutions for the equation $2x(8-7x) = -11$, rounded to two decimal places, are approximately $1.63$ and $-0.48$. This step-by-step calculation demonstrates how to apply the simplified quadratic formula to find the specific values of $x$ that satisfy the original equation. Rounding the solutions to a specified number of decimal places is often necessary for practical applications, ensuring that the answers are both accurate and useful.

Conclusion

In this article, we have systematically solved the quadratic equation $2x(8-7x) = -11$. We began by expanding and rearranging the equation into the standard quadratic form, which is a crucial step in solving any quadratic equation. This involved distributing the terms, moving all terms to one side of the equation, and ensuring the equation was in the form $ax^2 + bx + c = 0$. Next, we identified the coefficients $a$, $b$, and $c$ from the standard form, which are essential for applying the quadratic formula. We then substituted these coefficients into the quadratic formula:

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

This substitution is a critical step that requires careful attention to detail to avoid errors. Following the substitution, we simplified the expression by calculating the values under the square root and performing the necessary arithmetic operations. This simplification process broke down the complex formula into manageable steps, making it easier to find the solutions. We then calculated the two possible solutions for $x$ by considering both the addition and subtraction cases in the $\pm$ part of the formula. Finally, we rounded the solutions to two decimal places as requested, providing practical and accurate answers. The solutions we found were approximately $x_1 \approx 1.63$ and $x_2 \approx -0.48$. Understanding each step in this process is vital for mastering the solution of quadratic equations. The quadratic formula is a powerful tool in mathematics and has wide-ranging applications in various fields. By following this systematic approach, you can confidently solve quadratic equations and apply them to real-world problems. This comprehensive guide aims to provide you with the knowledge and skills necessary to tackle quadratic equations effectively.