Solving $2x^2 + X - 15 = 0$ By Completing The Square

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Solving $2x^2 + x - 15 = 0$ by Completing the Square

Hey everyone! Today, we're going to tackle the quadratic equation 2x2+xβˆ’15=02x^2 + x - 15 = 0 using the method of completing the square. This technique is super useful when factoring isn't straightforward or when you need to rewrite the quadratic in vertex form. So, let's dive right in and break down each step to make sure we understand exactly how to solve this.

Understanding the Method of Completing the Square

Before we jump into the specifics of our equation, let's quickly recap what completing the square actually means. In essence, it's a way of transforming a quadratic equation from its standard form, ax2+bx+c=0ax^2 + bx + c = 0, into a perfect square trinomial plus a constant. This form allows us to easily solve for xx by taking the square root. Completing the square involves algebraic manipulations to create a perfect square trinomial, which can then be factored into the form (x+h)2(x + h)^2 or (xβˆ’h)2(x - h)^2, where hh is a constant. This conversion simplifies the process of finding the roots (solutions) of the quadratic equation. The goal is to rewrite the equation in a form where we can isolate xx by taking the square root, making it a powerful technique when simple factorization isn't apparent or possible. By understanding this method, we can solve a broader range of quadratic equations efficiently.

Completing the square is particularly valuable because it provides a systematic approach to solving quadratic equations, even when the roots are irrational or complex. Unlike factoring, which relies on identifying integer or rational roots, completing the square works universally. Additionally, it lays the groundwork for deriving the quadratic formula, which is a generalized solution for any quadratic equation. The process involves several key steps:

  1. Divide by the leading coefficient: If the coefficient of x2x^2 (i.e., aa) is not 1, divide the entire equation by aa. This ensures that the quadratic term has a coefficient of 1, which is essential for completing the square.
  2. Move the constant term: Move the constant term (cc) to the right side of the equation. This isolates the x2x^2 and xx terms on one side, preparing the equation for the completion of the square.
  3. Complete the square: Take half of the coefficient of the xx term (i.e., b/2b/2), square it (i.e., (b/2)2(b/2)^2), and add it to both sides of the equation. This ensures that the left side becomes a perfect square trinomial.
  4. Factor the perfect square trinomial: Factor the left side of the equation into the form (x+b/2)2(x + b/2)^2 or (xβˆ’b/2)2(x - b/2)^2, depending on the sign of the xx term. This step simplifies the equation into a manageable form.
  5. Solve for x: Take the square root of both sides of the equation, remembering to include both the positive and negative roots. Then, solve for xx by isolating it on one side of the equation.

By following these steps, completing the square transforms a quadratic equation into a form that is easy to solve, providing a reliable method for finding the roots, regardless of their nature.

Step-by-Step Solution

Okay, let's get to it! Here’s how we’ll solve 2x2+xβˆ’15=02x^2 + x - 15 = 0 by completing the square.

Step 1: Divide by the Leading Coefficient

First, we need to make sure the coefficient of x2x^2 is 1. Currently, it's 2. So, we divide the entire equation by 2:

x2+12xβˆ’152=0x^2 + \frac{1}{2}x - \frac{15}{2} = 0

Dividing each term by the leading coefficient is a critical initial step in completing the square. This process normalizes the quadratic expression, making it easier to manipulate and transform into a perfect square trinomial. When the leading coefficient is not 1, the subsequent steps of completing the square become more complex and prone to errors. By dividing through by the leading coefficient, we ensure that the coefficient of the x2x^2 term is 1, which simplifies the arithmetic and algebraic manipulations required to complete the square. This step sets the stage for a smoother and more straightforward application of the completing the square method. It allows us to focus on the xx term and the constant term without the complication of an additional coefficient on the x2x^2 term. Furthermore, this normalization is essential for accurately determining the value needed to complete the square, as it directly impacts the term that must be added to both sides of the equation to create a perfect square trinomial.

Step 2: Move the Constant Term

Next, we move the constant term to the right side of the equation:

x2+12x=152x^2 + \frac{1}{2}x = \frac{15}{2}

Moving the constant term to the right side of the equation isolates the x2x^2 and xx terms on the left, which is a crucial step in preparing the equation for completing the square. This separation allows us to focus exclusively on the terms involving xx when we add the value needed to create a perfect square trinomial. By isolating these terms, we can clearly see what value must be added to both sides of the equation to complete the square effectively. This step streamlines the process by simplifying the expression and highlighting the necessary components for completing the square. It also ensures that the added term will correctly transform the left side into a perfect square, facilitating the subsequent factoring and solving steps. This isolation is essential for maintaining the balance and accuracy of the equation throughout the completion of the square process.

Step 3: Complete the Square

Now, we need to add a value to both sides to complete the square. To find this value, we take half of the coefficient of our xx term (which is 12\frac{1}{2}) and square it:

(12Γ·2)2=(14)2=116(\frac{1}{2} \div 2)^2 = (\frac{1}{4})^2 = \frac{1}{16}

So, we add 116\frac{1}{16} to both sides of the equation:

x2+12x+116=152+116x^2 + \frac{1}{2}x + \frac{1}{16} = \frac{15}{2} + \frac{1}{16}

Completing the square involves adding a specific value to both sides of the equation to create a perfect square trinomial on one side. This value is calculated by taking half of the coefficient of the xx term, squaring it, and then adding it to both sides of the equation. The rationale behind this process is that a perfect square trinomial can be factored into the form (x+a)2(x + a)^2 or (xβˆ’a)2(x - a)^2, which simplifies the equation and allows us to solve for xx more easily. By adding this calculated value, we ensure that the left side of the equation becomes a perfect square trinomial, making it factorable and setting the stage for the subsequent steps of completing the square. This step is essential for transforming the equation into a solvable form and is a core component of the completing the square method. The precision in calculating and adding this value is crucial for maintaining the balance of the equation and accurately creating the perfect square trinomial.

Step 4: Factor the Perfect Square Trinomial

Now, we factor the left side and simplify the right side:

(x+14)2=12016+116(x + \frac{1}{4})^2 = \frac{120}{16} + \frac{1}{16}

(x+14)2=12116(x + \frac{1}{4})^2 = \frac{121}{16}

Factoring the perfect square trinomial simplifies the equation into a manageable form, where the left side is expressed as a squared term. This step is crucial because it transforms the quadratic expression into a form that allows us to easily isolate xx by taking the square root of both sides. The factoring process involves recognizing that the perfect square trinomial can be written as (x+a)2(x + a)^2 or (xβˆ’a)2(x - a)^2, where aa is half of the coefficient of the xx term in the original quadratic expression. This factorization simplifies the equation and sets the stage for solving for xx. The accuracy of this step is vital, as an incorrect factorization will lead to incorrect solutions. By factoring the perfect square trinomial, we transform the equation into a form that is conducive to solving for the variable, making it a pivotal step in the completing the square method.

Step 5: Solve for x

Take the square root of both sides:

x+14=Β±12116x + \frac{1}{4} = \pm \sqrt{\frac{121}{16}}

x+14=Β±114x + \frac{1}{4} = \pm \frac{11}{4}

Now, isolate xx:

x=βˆ’14Β±114x = -\frac{1}{4} \pm \frac{11}{4}

So we have two possible solutions:

x1=βˆ’14+114=104=52x_1 = -\frac{1}{4} + \frac{11}{4} = \frac{10}{4} = \frac{5}{2}

x2=βˆ’14βˆ’114=βˆ’124=βˆ’3x_2 = -\frac{1}{4} - \frac{11}{4} = -\frac{12}{4} = -3

Solving for xx involves isolating the variable on one side of the equation to find its possible values. This process typically begins after factoring the perfect square trinomial and taking the square root of both sides of the equation. Taking the square root introduces both positive and negative roots, which must be considered to find all possible solutions for xx. After taking the square root, the equation is simplified, and xx can be isolated by performing basic algebraic operations such as addition, subtraction, multiplication, or division. Each step in this process must be performed accurately to ensure the correct values for xx are obtained. The final solutions represent the roots of the original quadratic equation, and verifying these solutions by substituting them back into the original equation can confirm their correctness. This entire process is essential for finding the values of xx that satisfy the equation and is a critical component of solving quadratic equations using the completing the square method.

Conclusion

Therefore, the solutions to the equation 2x2+xβˆ’15=02x^2 + x - 15 = 0 are x=52x = \frac{5}{2} and x=βˆ’3x = -3.

Completing the square might seem a bit involved at first, but with practice, it becomes a straightforward method to solve quadratic equations. Keep at it, and you’ll master it in no time!