Solving The Quadratic Equation X^2 + 6x = -8 A Step-by-Step Guide

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In the realm of mathematics, quadratic equations hold a significant place. They appear in various applications, from physics to engineering, and understanding how to solve them is crucial. In this article, we will delve into solving the specific quadratic equation x2+6x=−8x^2 + 6x = -8. We will explore the steps involved, the underlying concepts, and the different methods we can use to arrive at the solution.

Understanding Quadratic Equations

Before we dive into solving our equation, let's first understand what a quadratic equation is. A quadratic equation is a polynomial equation of the second degree. This means that the highest power of the variable in the equation is 2. The general form of a quadratic equation is:

ax2+bx+c=0ax^2 + bx + c = 0

where a, b, and c are constants, and a ≠ 0. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.

To effectively solve quadratic equations, it is essential to first recognize their standard form. The given equation, x2+6x=−8x^2 + 6x = -8, can be rearranged to fit this form, which is a crucial initial step. By adding 8 to both sides, we transform the equation into x2+6x+8=0x^2 + 6x + 8 = 0. This transformation not only makes the equation easier to solve but also allows us to readily identify the coefficients a, b, and c. In this case, a is 1, b is 6, and c is 8. Identifying these coefficients is vital because they are used in various methods for solving quadratic equations, such as factoring, completing the square, and applying the quadratic formula. Understanding the standard form ensures that we can apply these methods correctly and efficiently, leading to accurate solutions. Moreover, recognizing the standard form helps in visualizing the equation's graph as a parabola, which provides additional insights into the nature of the solutions.

Transforming the Equation

Our equation, x2+6x=−8x^2 + 6x = -8, is not yet in the standard form. To transform it, we need to move all terms to one side of the equation, leaving zero on the other side. To do this, we add 8 to both sides of the equation:

x2+6x+8=−8+8x^2 + 6x + 8 = -8 + 8

This simplifies to:

x2+6x+8=0x^2 + 6x + 8 = 0

Now, our equation is in the standard form, making it easier to solve.

Transforming a quadratic equation into its standard form is a foundational step in solving quadratic equations. This form, ax2+bx+c=0ax^2 + bx + c = 0, provides a clear structure that simplifies the application of various solution methods. In our example, the original equation x2+6x=−8x^2 + 6x = -8 is not immediately recognizable as a quadratic equation in standard form. The crucial step of adding 8 to both sides to obtain x2+6x+8=0x^2 + 6x + 8 = 0 transforms the equation into a format that is conducive to further analysis. This transformation not only reveals the coefficients a, b, and c, which are essential for methods like the quadratic formula and completing the square, but also sets the stage for factoring. Factoring becomes significantly easier when the equation is in standard form because it allows us to identify the specific combinations of factors that will lead to the solutions. Moreover, the standard form is essential for graphing the quadratic equation as a parabola, where the coefficients directly influence the shape and position of the parabola. Therefore, mastering the transformation into standard form is an indispensable skill for anyone seeking to solve quadratic equations effectively.

Methods for Solving Quadratic Equations

There are several methods for solving quadratic equations, each with its advantages and disadvantages. The most common methods include:

  1. Factoring: This method involves expressing the quadratic equation as a product of two linear factors.
  2. Completing the Square: This method involves manipulating the equation to create a perfect square trinomial.
  3. Quadratic Formula: This formula provides a direct solution for any quadratic equation.

We will explore each of these methods in detail as we solve our equation.

Choosing the right method for solving quadratic equations often depends on the specific characteristics of the equation. Factoring, for instance, is a powerful technique when the quadratic expression can be easily decomposed into two binomial factors. This method relies on the principle that if the product of two factors is zero, then at least one of the factors must be zero. For simple quadratic equations, factoring can be the quickest and most straightforward approach. However, not all quadratic equations are factorable over integers, which limits the applicability of this method. Completing the square is a more versatile technique that can be used to solve any quadratic equation. This method involves transforming the quadratic expression into a perfect square trinomial, which can then be easily solved by taking the square root. While completing the square is a reliable method, it can be more algebraically intensive than factoring, particularly when the coefficient of the x term is not an even number. The quadratic formula is the most general method for solving quadratic equations. It can be applied to any quadratic equation, regardless of whether it is factorable or not. The quadratic formula provides a direct solution for the roots of the equation, making it a valuable tool for solving complex quadratic equations. Each method offers a unique approach, and the choice of method often depends on the context and the solver's familiarity with the techniques.

Method 1: Factoring

Factoring involves breaking down the quadratic expression into two linear expressions whose product equals the original expression. To factor the equation x2+6x+8=0x^2 + 6x + 8 = 0, we need to find two numbers that add up to 6 (the coefficient of the x term) and multiply to 8 (the constant term).

The numbers 2 and 4 satisfy these conditions (2 + 4 = 6 and 2 * 4 = 8). Therefore, we can factor the equation as follows:

(x+2)(x+4)=0(x + 2)(x + 4) = 0

Now, for the product of two factors to be zero, at least one of the factors must be zero. So, we have two possibilities:

  1. x+2=0x + 2 = 0
  2. x+4=0x + 4 = 0

Solving these linear equations, we get:

  1. x=−2x = -2
  2. x=−4x = -4

Therefore, the solutions to the equation x2+6x+8=0x^2 + 6x + 8 = 0 are x = -2 and x = -4.

When solving quadratic equations, factoring stands out as an elegant and efficient method when applicable. The core principle behind factoring is to decompose the quadratic expression into two binomial factors. This method hinges on finding two numbers that satisfy specific conditions related to the coefficients of the quadratic equation. In the equation x2+6x+8=0x^2 + 6x + 8 = 0, the key is to identify two numbers that sum to 6 (the coefficient of the x term) and multiply to 8 (the constant term). Through careful consideration, we find that 2 and 4 meet these criteria. This allows us to rewrite the equation in factored form as (x+2)(x+4)=0(x + 2)(x + 4) = 0. The beauty of factoring lies in its simplicity: once the equation is factored, the solutions are easily obtained by setting each factor equal to zero. This leads to the equations x+2=0x + 2 = 0 and x+4=0x + 4 = 0, which yield the solutions x = -2 and x = -4. Factoring not only provides a direct path to the solutions but also enhances understanding of the equation's structure. It visually demonstrates how the roots of the equation are connected to the constant term and the coefficient of the x term. However, it's important to recognize that factoring is most effective when the roots are integers or simple fractions; for more complex roots, other methods like the quadratic formula or completing the square may be more suitable.

Method 2: Completing the Square

Completing the square is another method for solving quadratic equations. It involves manipulating the equation to create a perfect square trinomial on one side. To complete the square for the equation x2+6x+8=0x^2 + 6x + 8 = 0, we follow these steps:

  1. Move the constant term to the right side of the equation:

    x2+6x=−8x^2 + 6x = -8

  2. Take half of the coefficient of the x term (which is 6), square it (which is 9), and add it to both sides of the equation:

    x2+6x+9=−8+9x^2 + 6x + 9 = -8 + 9

  3. Rewrite the left side as a perfect square:

    (x+3)2=1(x + 3)^2 = 1

  4. Take the square root of both sides:

    x+3=±√1x + 3 = ±√1

  5. Solve for x:

    x=−3±1x = -3 ± 1

This gives us two solutions:

  1. x=−3+1=−2x = -3 + 1 = -2
  2. x=−3−1=−4x = -3 - 1 = -4

These are the same solutions we obtained using factoring.

Completing the square is a versatile method for solving quadratic equations, offering a systematic approach that is particularly useful when factoring is not straightforward. This technique involves transforming the quadratic equation into a form where one side is a perfect square trinomial. For the equation x2+6x+8=0x^2 + 6x + 8 = 0, the initial step is to isolate the constant term by moving it to the right side, resulting in x2+6x=−8x^2 + 6x = -8. The crucial part of completing the square is to add a value to both sides that makes the left side a perfect square. This value is determined by taking half of the coefficient of the x term (which is 6), squaring it (resulting in 9), and adding it to both sides, yielding x2+6x+9=−8+9x^2 + 6x + 9 = -8 + 9. The left side can then be rewritten as a perfect square, (x+3)2(x + 3)^2, and the right side simplifies to 1. Taking the square root of both sides gives x+3=±√1x + 3 = ±√1, which leads to x+3=±1x + 3 = ±1. Finally, solving for x involves subtracting 3 from both sides, giving us the two solutions: x=−3+1=−2x = -3 + 1 = -2 and x=−3−1=−4x = -3 - 1 = -4. Completing the square not only provides the solutions but also demonstrates the underlying structure of the quadratic equation, making it a valuable tool for deeper understanding. While it may seem more involved than factoring, completing the square is a powerful method that can be applied to any quadratic equation, regardless of whether it is factorable.

Method 3: Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations. It can be used to solve any quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0. The formula is given by:

x=−b±√(b2−4ac)2ax = \frac{-b ± √(b^2 - 4ac)}{2a}

For our equation, x2+6x+8=0x^2 + 6x + 8 = 0, we have a = 1, b = 6, and c = 8. Substituting these values into the quadratic formula, we get:

x=−6±√(62−4∗1∗8)2∗1x = \frac{-6 ± √(6^2 - 4 * 1 * 8)}{2 * 1}

Simplifying, we get:

x=−6±√(36−32)2x = \frac{-6 ± √(36 - 32)}{2}

x=−6±√42x = \frac{-6 ± √4}{2}

x=−6±22x = \frac{-6 ± 2}{2}

This gives us two solutions:

  1. x=−6+22=−42=−2x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2
  2. x=−6−22=−82=−4x = \frac{-6 - 2}{2} = \frac{-8}{2} = -4

Again, we obtain the same solutions: x = -2 and x = -4.

Employing the quadratic formula is the most general and foolproof method for solving quadratic equations, capable of tackling any equation of the form ax2+bx+c=0ax^2 + bx + c = 0. The formula, x=−b±√(b2−4ac)2ax = \frac{-b ± √(b^2 - 4ac)}{2a}, is a direct pathway to the solutions, irrespective of the equation's complexity. In our case, the equation x2+6x+8=0x^2 + 6x + 8 = 0 has coefficients a = 1, b = 6, and c = 8. Substituting these values into the quadratic formula, we get x=−6±√(62−4∗1∗8)2∗1x = \frac{-6 ± √(6^2 - 4 * 1 * 8)}{2 * 1}. The subsequent simplification involves evaluating the discriminant, b2−4acb^2 - 4ac, which in this case is 62−4∗1∗8=36−32=46^2 - 4 * 1 * 8 = 36 - 32 = 4. This leads to x=−6±√42x = \frac{-6 ± √4}{2}, which further simplifies to x=−6±22x = \frac{-6 ± 2}{2}. The ± sign indicates that there are two potential solutions: one where we add 2 and one where we subtract 2. The first solution is x=−6+22=−42=−2x = \frac{-6 + 2}{2} = \frac{-4}{2} = -2, and the second solution is x=−6−22=−82=−4x = \frac{-6 - 2}{2} = \frac{-8}{2} = -4. These solutions align perfectly with those obtained through factoring and completing the square, reinforcing the consistency and reliability of the quadratic formula. The formula is particularly invaluable when dealing with equations that have irrational or complex roots, where other methods may not be as straightforward.

Conclusion

In this article, we successfully solved the quadratic equation x2+6x=−8x^2 + 6x = -8 using three different methods: factoring, completing the square, and the quadratic formula. All three methods yielded the same solutions, x = -2 and x = -4. This demonstrates the versatility of quadratic equations and the various approaches we can take to solve them. Understanding these methods is crucial for anyone studying mathematics and its applications.

Solving the quadratic equation x2+6x=−8x^2 + 6x = -8 exemplifies the diverse techniques available in solving quadratic equations. Throughout this exploration, we applied factoring, completing the square, and the quadratic formula, each method providing a unique path to the solutions x = -2 and x = -4. The consistency of these results underscores the robustness of quadratic equation-solving methods. Factoring allowed us to express the quadratic as a product of two binomials, directly revealing the roots. Completing the square transformed the equation into a perfect square trinomial, offering a systematic algebraic approach. The quadratic formula provided a universal solution, applicable to any quadratic equation, regardless of its factorability. Mastering these methods is essential for a comprehensive understanding of algebra and its applications. Quadratic equations appear in various fields, from physics and engineering to economics and computer science. The ability to solve them efficiently and accurately is a valuable skill that enhances problem-solving capabilities. This article not only provides a step-by-step guide to solving a specific quadratic equation but also highlights the broader importance of understanding and applying different solution methods in mathematics.