Completing The Square Find The Number To Add To X^2 + 8x = 4

by ADMIN 61 views

Completing the square is a fundamental technique in algebra for solving quadratic equations, transforming them into a more manageable form. This method allows us to rewrite a quadratic equation in the form (x+a)2=b(x + a)^2 = b, from which we can easily find the solutions for x. In this article, we will delve into the process of completing the square, specifically focusing on the equation x2+8x=4x^2 + 8x = 4. We aim to identify the number that needs to be added to both sides of the equation to achieve the perfect square trinomial on the left-hand side. This involves understanding the relationship between the coefficients of the quadratic and linear terms and how they dictate the constant term required to complete the square.

Understanding the Process of Completing the Square

The core idea behind completing the square lies in manipulating a quadratic expression to create a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial, such as (x+a)2(x + a)^2 or (x−a)2(x - a)^2. When we expand these binomial squares, we get:

(x+a)2=x2+2ax+a2(x + a)^2 = x^2 + 2ax + a^2

(x−a)2=x2−2ax+a2(x - a)^2 = x^2 - 2ax + a^2

Notice the pattern: the constant term, a2a^2, is the square of half the coefficient of the linear term (the term with x). This observation is key to completing the square. In our given equation, x2+8x=4x^2 + 8x = 4, we want to transform the left-hand side into a perfect square trinomial. To do this, we need to find the appropriate constant term to add.

To effectively complete the square, first consider the given equation: x2+8x=4x^2 + 8x = 4. The left-hand side currently consists of a quadratic term (x2x^2) and a linear term (8x8x). To complete the square, we need to add a constant term that will make the left-hand side a perfect square trinomial. This constant term must be such that the resulting trinomial can be factored into the form (x+a)2(x + a)^2.

To find this constant term, we focus on the coefficient of the linear term, which is 8 in this case. According to the perfect square trinomial pattern, the constant term should be the square of half this coefficient. Therefore, we take half of 8, which is 4, and then square it: 42=164^2 = 16. This means that 16 is the constant term we need to add to both sides of the equation to complete the square. Adding 16 to both sides gives us:

x2+8x+16=4+16x^2 + 8x + 16 = 4 + 16

Now, the left-hand side, x2+8x+16x^2 + 8x + 16, is a perfect square trinomial that can be factored into (x+4)2(x + 4)^2. The right-hand side simplifies to 20, so our equation becomes:

(x+4)2=20(x + 4)^2 = 20

This transformation is the essence of completing the square. By adding the correct constant term, we have rewritten the quadratic equation in a form that allows us to easily solve for x by taking the square root of both sides.

Step-by-Step Solution for x2+8x=4x^2 + 8x = 4

Let's break down the process step-by-step to solidify the concept:

  1. Identify the coefficient of the linear term: In the equation x2+8x=4x^2 + 8x = 4, the coefficient of the linear term (x) is 8.

  2. Divide the coefficient by 2: Divide 8 by 2, which gives us 4.

  3. Square the result: Square 4, which gives us 42=164^2 = 16.

  4. Add this number to both sides of the equation: Add 16 to both sides of the original equation:

    x2+8x+16=4+16x^2 + 8x + 16 = 4 + 16

  5. Factor the left-hand side as a perfect square trinomial: The left-hand side now factors into (x+4)2(x + 4)^2.

  6. Simplify the right-hand side: The right-hand side simplifies to 20.

The equation now looks like this:

(x+4)2=20(x + 4)^2 = 20

This form allows us to easily solve for x by taking the square root of both sides. The number we added to both sides, 16, is the key to completing the square in this particular equation. Completing the square is a versatile method that provides a systematic way to solve any quadratic equation, regardless of whether it can be easily factored using other techniques.

Why Completing the Square Works

The reason completing the square works so effectively is rooted in the algebraic identity of a perfect square trinomial. As mentioned earlier, a perfect square trinomial can be expressed in the form (x+a)2(x + a)^2 or (x−a)2(x - a)^2. Expanding (x+a)2(x + a)^2, we get:

(x+a)2=x2+2ax+a2(x + a)^2 = x^2 + 2ax + a^2

In the equation x2+8x=4x^2 + 8x = 4, we can see that the left-hand side has the form x2+2axx^2 + 2ax, where 2a=82a = 8. This implies that a=4a = 4. To complete the square, we need to add a2a^2 to both sides of the equation. Since a=4a = 4, we have a2=42=16a^2 = 4^2 = 16. Adding 16 to both sides allows us to rewrite the left-hand side as a perfect square trinomial:

x2+8x+16=(x+4)2x^2 + 8x + 16 = (x + 4)^2

The process of completing the square essentially transforms the original quadratic expression into a form that fits the perfect square trinomial identity. This transformation makes it straightforward to isolate x and find the solutions to the equation. The visual representation of completing the square, often demonstrated using geometric shapes, further illustrates this concept. Imagine a square with side length x, representing x2x^2, and two rectangles with sides x and 4, representing 4x4x each (which combine to give 8x8x). To complete the square, we need to add a smaller square with side length 4, representing 42=164^2 = 16. This visual approach underscores the algebraic manipulation we perform when completing the square.

Common Mistakes to Avoid

When completing the square, there are several common mistakes that students often make. Being aware of these pitfalls can help you avoid them and master the technique more effectively. One of the most frequent errors is forgetting to add the constant term to both sides of the equation. Remember, to maintain the equality, any operation performed on one side must also be performed on the other side. In the case of x2+8x=4x^2 + 8x = 4, if you add 16 only to the left-hand side, you change the equation and will not arrive at the correct solution.

Another common mistake is incorrectly calculating the constant term to add. It is crucial to remember that you need to take half of the coefficient of the linear term and then square it. For example, if the equation is x2+10x=5x^2 + 10x = 5, students might mistakenly square the entire coefficient (10) instead of halving it first. The correct constant term to add would be (10/2)2=52=25(10/2)^2 = 5^2 = 25, not 102=10010^2 = 100.

Furthermore, students sometimes struggle with factoring the perfect square trinomial once the constant term has been added. It's important to recognize the pattern of a perfect square trinomial, which can be factored into the form (x+a)2(x + a)^2 or (x−a)2(x - a)^2. For instance, x2+8x+16x^2 + 8x + 16 factors into (x+4)2(x + 4)^2 because 16 is the square of 4, and 8 is twice 4. Consistent practice with factoring perfect square trinomials will help solidify this skill.

Lastly, be mindful of equations where the coefficient of the x2x^2 term is not 1. In such cases, you must first divide the entire equation by this coefficient before completing the square. For example, if the equation is 2x2+8x=102x^2 + 8x = 10, you would first divide every term by 2 to get x2+4x=5x^2 + 4x = 5, and then proceed with completing the square.

By avoiding these common mistakes, you can confidently and accurately complete the square to solve quadratic equations.

Applying Completing the Square in Different Scenarios

The technique of completing the square is not just a method for solving quadratic equations; it's a versatile tool with applications in various areas of mathematics. One significant application is in transforming the general form of a quadratic equation into the vertex form. The vertex form of a quadratic equation is given by:

y=a(x−h)2+ky = a(x - h)^2 + k

where (h, k) represents the vertex of the parabola. Converting a quadratic equation from the standard form (y=ax2+bx+cy = ax^2 + bx + c) to the vertex form allows us to easily identify the vertex, axis of symmetry, and other key features of the parabola.

To convert to vertex form, we complete the square on the quadratic expression. For example, consider the equation y=x2+6x+5y = x^2 + 6x + 5. To convert this to vertex form, we focus on the x2+6xx^2 + 6x part and complete the square. Half of 6 is 3, and 323^2 is 9, so we add and subtract 9 within the equation:

y=(x2+6x+9)+5−9y = (x^2 + 6x + 9) + 5 - 9

Now, we can factor the perfect square trinomial:

y=(x+3)2−4y = (x + 3)^2 - 4

This is the vertex form of the equation, and we can immediately see that the vertex of the parabola is (-3, -4). This application highlights the power of completing the square in analyzing quadratic functions and their graphs.

Another important application of completing the square is in deriving the quadratic formula. The quadratic formula provides a general solution for any quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0:

x=−b±b2−4ac2ax = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}

The quadratic formula is derived by completing the square on the general quadratic equation. This derivation demonstrates the fundamental nature of completing the square as a problem-solving tool in algebra. Furthermore, completing the square is used in calculus, particularly in integration techniques, and in various areas of physics and engineering where quadratic equations frequently arise.

Conclusion

In conclusion, to complete the square for the equation x2+8x=4x^2 + 8x = 4, the number that should be added to both sides is 16. This process transforms the left-hand side into a perfect square trinomial, (x+4)2(x + 4)^2, allowing us to rewrite the equation as (x+4)2=20(x + 4)^2 = 20. Completing the square is a powerful algebraic technique with wide-ranging applications beyond solving quadratic equations. By mastering this method, you gain a valuable tool for tackling a variety of mathematical problems and a deeper understanding of quadratic expressions and their properties.

Therefore, the correct answer is C. 16. Understanding the mechanics of completing the square is crucial for success in algebra and higher-level mathematics. By practicing and applying this technique in various contexts, you will develop a strong foundation for problem-solving and mathematical reasoning.