Solving Quadratic Equations Completing The Square Method

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Understanding how to solve quadratic equations is a fundamental skill in algebra. One powerful method for tackling these equations is completing the square. This technique transforms a quadratic equation into a perfect square trinomial, making it easier to isolate the variable and find the solutions. In this article, we will delve into the process of completing the square, step by step, and illustrate it with an example. We will analyze the specific steps needed to solve the quadratic equation 5x2+20x−7=05x^2 + 20x - 7 = 0 by completing the square, and identify the correct options from a given set of steps.

Understanding Quadratic Equations

A quadratic equation is a polynomial equation of the second degree. 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 solutions to a quadratic equation are the values of x that satisfy the equation. These solutions are also known as the roots or zeros of the quadratic equation. There are several methods to solve quadratic equations, including factoring, using the quadratic formula, and completing the square. Each method has its advantages, and the choice of method often depends on the specific equation at hand. Completing the square is particularly useful because it can be applied to any quadratic equation, regardless of whether it can be easily factored.

The Power of Completing the Square

Completing the square is a technique that rewrites a quadratic expression in the form of a perfect square trinomial plus a constant. This method is incredibly versatile because it allows us to transform any quadratic equation into a form where we can easily isolate the variable x. The underlying principle is to manipulate the equation so that one side becomes a perfect square, which can then be easily solved by taking the square root. This technique is not only useful for solving equations but also for converting quadratic equations into vertex form, which helps in graphing parabolas and finding their minimum or maximum values. By mastering completing the square, you gain a powerful tool that simplifies many algebraic problems and provides a deeper understanding of quadratic functions.

Steps to Completing the Square

To effectively use the method of completing the square, it is crucial to follow a structured approach. Here’s a detailed breakdown of the steps involved:

  1. Divide by the Leading Coefficient: If the coefficient of x2x^2 (i.e., a) is not 1, divide the entire equation by a. This ensures that the coefficient of x2x^2 becomes 1, which is necessary for the subsequent steps. This step simplifies the process and sets the stage for creating a perfect square trinomial. For example, if you have 2x2+8x−10=02x^2 + 8x - 10 = 0, you would divide the entire equation by 2, resulting in x2+4x−5=0x^2 + 4x - 5 = 0.

  2. Move the Constant Term: Move the constant term (c) to the right side of the equation. This isolates the terms containing x on one side, preparing the equation for the completion of the square. By isolating these terms, we can focus on creating the perfect square trinomial without the distraction of the constant term. For instance, in the equation x2+4x−5=0x^2 + 4x - 5 = 0, you would add 5 to both sides to get x2+4x=5x^2 + 4x = 5.

  3. Complete the Square: Take half of the coefficient of the x term (i.e., b), square it, and add the result to both sides of the equation. This is the core of the method, as it creates a perfect square trinomial on the left side. The value you add is (b2)2(\frac{b}{2})^2. Adding the same value to both sides ensures that the equation remains balanced. For example, if the equation is x2+4x=5x^2 + 4x = 5, half of 4 is 2, and 222^2 is 4. So, you would add 4 to both sides, resulting in x2+4x+4=5+4x^2 + 4x + 4 = 5 + 4.

  4. Factor the Perfect Square Trinomial: The left side of the equation should now be a perfect square trinomial. Factor it into the form (x+b2)2(x + \frac{b}{2})^2 or (x−b2)2(x - \frac{b}{2})^2, depending on the sign of the x term. This step simplifies the equation and makes it clear that we have created a perfect square. For example, x2+4x+4x^2 + 4x + 4 can be factored into (x+2)2(x + 2)^2, so the equation becomes (x+2)2=9(x + 2)^2 = 9.

  5. Take the Square Root: Take the square root of both sides of the equation. Remember to consider both positive and negative square roots. This step undoes the square, allowing us to isolate the variable x. For instance, taking the square root of both sides of (x+2)2=9(x + 2)^2 = 9 gives x+2=±3x + 2 = \pm 3.

  6. Solve for x: Solve the resulting equations for x. This usually involves isolating x by performing basic algebraic operations. This step gives the solutions to the original quadratic equation. For example, from x+2=±3x + 2 = \pm 3, we get two equations: x+2=3x + 2 = 3 and x+2=−3x + 2 = -3. Solving these gives x=1x = 1 and x=−5x = -5.

Applying the Steps to 5x2+20x−7=05x^2 + 20x - 7 = 0

Now, let's apply these steps to the given equation: 5x2+20x−7=05x^2 + 20x - 7 = 0. This will provide a concrete example of how to use the method of completing the square to solve a quadratic equation.

  1. Divide by the Leading Coefficient: The leading coefficient is 5, so we divide the entire equation by 5:

x2+4x−75=0x^2 + 4x - \frac{7}{5} = 0

This step is crucial because it simplifies the equation and makes it easier to complete the square in the subsequent steps. By dividing by the leading coefficient, we ensure that the coefficient of x2x^2 is 1, which is a prerequisite for creating a perfect square trinomial.

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

x2+4x=75x^2 + 4x = \frac{7}{5}

By isolating the terms containing x on one side, we set the stage for completing the square. This step allows us to focus on creating a perfect square trinomial without the distraction of the constant term. The equation is now in a form that is more amenable to the completion of the square process.

  1. Complete the Square: Take half of the coefficient of the x term (which is 4), square it, and add the result to both sides. Half of 4 is 2, and 222^2 is 4. So, we add 4 to both sides:

x2+4x+4=75+4x^2 + 4x + 4 = \frac{7}{5} + 4

Adding 4 to both sides ensures that the equation remains balanced while creating the perfect square trinomial on the left side. This step is the heart of the completing the square method, transforming the equation into a more manageable form.

  1. Factor the Perfect Square Trinomial: Factor the left side as a perfect square trinomial:

(x+2)2=75+4(x + 2)^2 = \frac{7}{5} + 4

(x+2)2=75+205(x + 2)^2 = \frac{7}{5} + \frac{20}{5}

(x+2)2=275(x + 2)^2 = \frac{27}{5}

This step simplifies the equation, making it clear that we have successfully created a perfect square on one side. The factored form allows us to easily take the square root in the next step, progressing towards the solution of the equation.

  1. Take the Square Root: Take the square root of both sides:

x+2=±275x + 2 = \pm \sqrt{\frac{27}{5}}

Remembering to consider both positive and negative square roots is crucial, as quadratic equations typically have two solutions. This step undoes the square, bringing us closer to isolating the variable x.

  1. Solve for x: Subtract 2 from both sides to solve for x:

x=−2±275x = -2 \pm \sqrt{\frac{27}{5}}

This gives us the two solutions for the quadratic equation. These solutions are the values of x that satisfy the original equation, and they represent the points where the parabola intersects the x-axis.

Analyzing the Given Options

Now, let's revisit the options provided in the question and determine which steps Sergey could use to solve the quadratic equation 5x2+20x−7=05x^2 + 20x - 7 = 0 by completing the square.

  • Option A: 5(x2+4x+4)=−7+205(x^2 + 4x + 4) = -7 + 20

    This step is derived from the process of completing the square. Starting from 5x2+20x=75x^2 + 20x = 7, we factor out 5 to get 5(x2+4x)=75(x^2 + 4x) = 7. To complete the square, we need to add (42)2=4(\frac{4}{2})^2 = 4 inside the parenthesis. This means we are actually adding 5×4=205 \times 4 = 20 to the left side, so we must add 20 to the right side as well. Thus, 5(x2+4x+4)=7+205(x^2 + 4x + 4) = 7 + 20 which simplifies to 5(x2+4x+4)=275(x^2 + 4x + 4) = 27. So, Option A is a correct step.

  • Option B: x+2=±275x + 2 = \pm \sqrt{\frac{27}{5}}

    This step follows logically from the completed square form. After completing the square, we have (x+2)2=275(x + 2)^2 = \frac{27}{5}. Taking the square root of both sides gives x+2=±275x + 2 = \pm \sqrt{\frac{27}{5}}. Thus, Option B is a correct step.

  • Option C: 5(x2+4x)=75(x^2 + 4x) = 7

    This is the initial step in preparing the equation for completing the square. By moving the constant term to the right side, we isolate the x terms, which is essential for completing the square. Starting with 5x2+20x−7=05x^2 + 20x - 7 = 0, we add 7 to both sides to get 5x2+20x=75x^2 + 20x = 7. Factoring out 5 from the left side gives 5(x2+4x)=75(x^2 + 4x) = 7. So, Option C is a correct step.

Conclusion

In summary, the correct steps Sergey could use to solve the quadratic equation 5x2+20x−7=05x^2 + 20x - 7 = 0 by completing the square are:

  • 5(x2+4x)=75(x^2 + 4x) = 7
  • 5(x2+4x+4)=−7+205(x^2 + 4x + 4) = -7 + 20
  • x+2=±275x + 2 = \pm \sqrt{\frac{27}{5}}

Completing the square is a versatile method for solving quadratic equations, and understanding the steps involved is crucial for mastering this technique. By following a structured approach, you can transform any quadratic equation into a solvable form, enhancing your problem-solving skills in algebra. This method not only helps in finding solutions but also provides a deeper understanding of the structure and properties of quadratic equations.