Solving Quadratic Equations By Completing The Square

In mathematics, quadratic equations play a pivotal role, appearing in various fields like physics, engineering, and economics. A quadratic equation is a polynomial equation of the second degree, generally represented in the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. Among the several methods to solve quadratic equations, completing the square stands out as a powerful technique. This method not only helps in finding the solutions (roots) of the equation but also provides a deeper understanding of the structure of quadratic expressions. In this comprehensive guide, we will delve into the method of completing the square, illustrating its application through a series of examples. We will tackle equations with varying complexities, ensuring a thorough grasp of the technique.

Understanding the Method of Completing the Square

At its core, completing the square is a technique used to rewrite a quadratic expression in the form (x + p)² + q, where p and q are constants. This form is particularly useful because it allows us to easily identify the vertex of the parabola represented by the quadratic equation and, more importantly, to solve for the roots of the equation. The method involves manipulating the quadratic expression by adding and subtracting a specific constant, which transforms part of the expression into a perfect square trinomial. This perfect square trinomial can then be factored into the square of a binomial, leading us closer to the solution.

The beauty of completing the square lies in its ability to convert any quadratic equation into a form that can be easily solved by taking the square root. This method is especially valuable when the quadratic equation cannot be easily factored. Furthermore, understanding completing the square provides a solid foundation for deriving the quadratic formula, a universal solution for any quadratic equation. The process might seem a bit intricate at first, but with practice, it becomes a straightforward and elegant way to solve quadratic equations.

To effectively use the completing the square method, it's essential to grasp the concept of a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. For example, x² + 6x + 9 is a perfect square trinomial because it can be factored into (x + 3)². The key to completing the square is to manipulate the given quadratic equation to create such a trinomial. This involves identifying the coefficient of the x term, halving it, squaring the result, and then adding and subtracting this value within the equation. This process doesn't change the value of the equation but rearranges it into a more solvable form. The ability to recognize and create perfect square trinomials is the cornerstone of mastering the completing the square method, making it an indispensable tool in solving quadratic equations.

Example 1: Solving x² + 10x + 2 = 0 by Completing the Square

Let's begin with our first example: x² + 10x + 2 = 0. This equation is a classic illustration of how completing the square can be applied. Our first step is to isolate the constant term on the right side of the equation. This means subtracting 2 from both sides, giving us x² + 10x = -2. This rearrangement sets the stage for creating our perfect square trinomial. The next critical step involves identifying the coefficient of the x term, which in this case is 10. We then halve this coefficient (10 / 2 = 5) and square the result (5² = 25). This value, 25, is the number we need to add to both sides of the equation to complete the square.

Adding 25 to both sides of x² + 10x = -2 gives us x² + 10x + 25 = -2 + 25, which simplifies to x² + 10x + 25 = 23. Notice that the left side of the equation is now a perfect square trinomial. It can be factored into (x + 5)². So, our equation becomes (x + 5)² = 23. This transformation is the heart of the completing the square method – we've successfully rewritten the quadratic expression in a form that's much easier to solve.

Now that we have (x + 5)² = 23, the next step is to take the square root of both sides. This gives us x + 5 = ±√23. Remember to include both the positive and negative square roots, as both values will satisfy the equation. Finally, we isolate x by subtracting 5 from both sides, resulting in x = -5 ± √23. These are the two solutions to the quadratic equation x² + 10x + 2 = 0. We have successfully solved the equation by completing the square, demonstrating the power and elegance of this method. This example highlights the step-by-step process, from isolating the constant term to taking the square root, providing a clear roadmap for tackling similar problems.

Example 2: Solving 2y² - 3 = 5y by Completing the Square

Our second example, 2y² - 3 = 5y, introduces a slight twist as the coefficient of the y² term is not 1. To effectively apply completing the square, we first need to ensure that the coefficient of the squared term is 1. Let's start by rearranging the equation to bring all terms to one side, resulting in 2y² - 5y - 3 = 0. The next crucial step is to divide the entire equation by the coefficient of the y² term, which is 2. This gives us y² - (5/2)y - (3/2) = 0. Now, the equation is in the standard form required for completing the square.

Next, we isolate the constant term by adding (3/2) to both sides, which yields y² - (5/2)y = 3/2. Now, we focus on the coefficient of the y term, which is (-5/2). We halve this coefficient, getting (-5/4), and then square it, resulting in 25/16. This value is what we need to add to both sides of the equation to complete the square. Adding 25/16 to both sides gives us y² - (5/2)y + 25/16 = 3/2 + 25/16. The left side of the equation is now a perfect square trinomial, and the right side can be simplified.

The left side, y² - (5/2)y + 25/16, can be factored into (y - 5/4)². The right side, 3/2 + 25/16, simplifies to 49/16. Thus, our equation becomes (y - 5/4)² = 49/16. Now, we take the square root of both sides, remembering to include both the positive and negative roots, which gives us y - 5/4 = ±√(49/16). This simplifies to y - 5/4 = ±7/4. Finally, we isolate y by adding 5/4 to both sides, resulting in y = 5/4 ± 7/4. This gives us two solutions: y = 3 and y = -1/2. This example demonstrates how to handle quadratic equations where the coefficient of the squared term is not 1, highlighting the necessary adjustments to the completing the square method.

Example 3: Solving b² - 2b - 15 = 0 by Completing the Square

Our third example, b² - 2b - 15 = 0, presents a straightforward application of completing the square. The equation is already in a suitable form, with the coefficient of the b² term being 1. We begin by isolating the constant term, which means adding 15 to both sides of the equation. This gives us b² - 2b = 15. Now, we focus on the coefficient of the b term, which is -2. We halve this coefficient (-2 / 2 = -1) and square the result ((-1)² = 1). This value, 1, is what we need to add to both sides of the equation to complete the square.

Adding 1 to both sides of b² - 2b = 15 gives us b² - 2b + 1 = 15 + 1, which simplifies to b² - 2b + 1 = 16. The left side of the equation is now a perfect square trinomial, which can be factored into (b - 1)². So, our equation becomes (b - 1)² = 16. This is a significant step, as we've transformed the quadratic expression into a more manageable form.

Next, we take the square root of both sides of the equation, remembering to consider both positive and negative roots. This gives us b - 1 = ±√16, which simplifies to b - 1 = ±4. To solve for b, we add 1 to both sides, resulting in b = 1 ± 4. This gives us two solutions: b = 5 and b = -3. This example neatly illustrates the process of completing the square when the equation is already in a relatively simple form, reinforcing the key steps of isolating the constant term, finding the value to complete the square, and solving for the variable.

Example 4: Solving 3n² + 24n = -48 by Completing the Square

In our fourth example, 3n² + 24n = -48, we encounter another scenario where the coefficient of the squared term is not 1. As we learned in Example 2, the first step is to make this coefficient 1. We achieve this by dividing the entire equation by 3, which gives us n² + 8n = -16. This step is crucial for setting up the equation for completing the square.

Now, we focus on the left side of the equation, n² + 8n. The coefficient of the n term is 8. We halve this coefficient (8 / 2 = 4) and square the result (4² = 16). This value, 16, is exactly what we need to add to both sides of the equation to complete the square. Adding 16 to both sides gives us n² + 8n + 16 = -16 + 16, which simplifies to n² + 8n + 16 = 0. Notice that the right side of the equation is now 0, which will simplify our solution process.

The left side of the equation, n² + 8n + 16, is a perfect square trinomial and can be factored into (n + 4)². So, our equation becomes (n + 4)² = 0. To solve for n, we take the square root of both sides, which gives us n + 4 = 0. Subtracting 4 from both sides, we find that n = -4. In this case, we have a single, repeated solution. This example demonstrates how completing the square works when the quadratic equation has exactly one real solution, highlighting the importance of careful algebraic manipulation.

Example 5: Solving 2x² + 4x + 1 = 0 by Completing the Square

Our final example, 2x² + 4x + 1 = 0, presents a comprehensive challenge that encapsulates many of the techniques we've discussed. As before, we start by ensuring that the coefficient of the x² term is 1. To do this, we divide the entire equation by 2, which gives us x² + 2x + 1/2 = 0. This is a crucial first step in preparing the equation for completing the square.

Next, we isolate the constant term by subtracting 1/2 from both sides, resulting in x² + 2x = -1/2. Now, we focus on the coefficient of the x term, which is 2. We halve this coefficient (2 / 2 = 1) and square the result (1² = 1). This value, 1, is what we need to add to both sides of the equation to complete the square. Adding 1 to both sides gives us x² + 2x + 1 = -1/2 + 1, which simplifies to x² + 2x + 1 = 1/2.

The left side of the equation, x² + 2x + 1, is a perfect square trinomial and can be factored into (x + 1)². So, our equation becomes (x + 1)² = 1/2. To solve for x, we take the square root of both sides, remembering to include both positive and negative roots. This gives us x + 1 = ±√(1/2). We can rewrite √(1/2) as √1/√2, which simplifies to 1/√2. Rationalizing the denominator, we multiply the numerator and denominator by √2, resulting in √2/2. So, our equation becomes x + 1 = ±√2/2.

Finally, we isolate x by subtracting 1 from both sides, resulting in x = -1 ± √2/2. This gives us two solutions: x = -1 + √2/2 and x = -1 - √2/2. This example demonstrates the entire process of completing the square, including dealing with non-integer coefficients and rationalizing denominators, providing a robust understanding of the method.

Conclusion

Completing the square is a versatile and powerful technique for solving quadratic equations. As demonstrated through these five examples, the method can be applied to a wide range of quadratic equations, regardless of the coefficients involved. From simple equations to those requiring more intricate manipulation, completing the square provides a systematic approach to finding solutions. The key lies in understanding the underlying principles: creating a perfect square trinomial and then solving for the variable. Mastering this technique not only enhances your ability to solve quadratic equations but also deepens your understanding of algebraic manipulations and the structure of quadratic expressions. Whether you're a student learning algebra or someone looking to refresh your math skills, completing the square is an essential tool in your mathematical toolkit.