Problem 24 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation by completing the square. $$ 10 x-5=4 x^{2} $$

Step-by-Step Solution

Answer

The solutions to the equation \(4x^2 - 10x + 5 = 0\) by completing the square are \(x_1 = \frac{3}{2}\) and \(x_2 = 1\).

Step 1: Rewrite the equation

First, let's rewrite the equation as a quadratic equation in standard form: \( 4x^2 - 10x + 5 = 0 \)

Step 2: Complete the square

Next, complete the square to make solving easier: 1. Divide all terms by the coefficient of the x² term (4). \( x^2 - \frac{5}{2}x + \frac{5}{4} = 0 \) 2. Add and subtract \(\left(\frac{b}{2a}\right)^2\) inside the parentheses. Here, \(a = 1\) and \(b = -\frac{5}{2}\). So, \( \left(\frac{b}{2a}\right)^2 = \left(\frac{-5/2}{2(1)}\right)^2 = \left(\frac{-5}{4}\right)^2 = \frac{25}{16} \) 3. Rewrite the equation with the completed square. \( \left(x^2 - \frac{5}{2}x + \frac{25}{16}\right) - \frac{25}{16} + \frac{5}{4} = 0 \) 4. Factor the completed square and simplify. \( \left(x - \frac{5}{4}\right)^2 - \frac{1}{16} = 0 \)

Step 3: Solve for x

Now, solve for x by isolating the square term and taking the square root: 1. Add \(\frac{1}{16}\) to both sides of the equation. \( \left(x - \frac{5}{4}\right)^2 = \frac{1}{16} \) 2. Take the square root of both sides of the equation. \( x - \frac{5}{4} = \pm \frac{1}{4} \) 3. Solve for x. \( x = \frac{5}{4} \pm \frac{1}{4} \) So, the two possible solutions for x are: \[ x_1 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2} \] \[ x_2 = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1 \] The solutions to the equation are \(x_1 = \frac{3}{2}\) and \(x_2 = 1\).

Key Concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial equation that typically appears in the form:\[ax^2 + bx + c = 0\]where \(a, b,\) and \(c\) are constants, and \(a eq 0\). Quadratic equations can appear complex at first, but they follow consistent patterns that can be used for solving them. Understanding these patterns is key to mastering this type of problem.

Quadratic equations can present in several forms:

• Factored form: \((px + q)(rx + s) = 0\)
• Vertex form: \(y = a(x-h)^2 + k\)
• Standard form: \(ax^2 + bx + c = 0\)

Each form allows different techniques or perspectives for solving or interpreting the equation. In our original exercise, the equation was converted into the standard form, \(4x^2 - 10x + 5 = 0\), before proceeding to solve it by completing the square. This initial step helps clarify the equation structure and aids in using algebraic methods effectively.

Algebraic Manipulation

Algebraic manipulation involves reshaping an equation or expression to reveal more information or to solve it. It includes a series of operations such as addition, subtraction, multiplication, division, and factoring, aimed at simplifying problems and making equations solvable. In the context of our exercise, algebraic manipulation was used in several ways.

One critical algebraic manipulation technique is transferring terms to different sides of an equation to isolate variables or constants. For instance, our step-by-step solution began with moving terms around,transforming the linear form \(10x - 5 = 4x^2\) into a more manageable quadratic form \(4x^2 - 10x + 5 = 0\).

Additionally, dividing terms by a common factor, as done with the coefficient of \(x^2\) (which was 4 in our exercise), optimizes the manipulation process by simplifying expressions.Understanding these manipulative steps is essential for solving higher-level algebraic equations and for preparing numeric equations for further techniques like completing the square.

Square Root Method

The square root method is a technique used to solve quadratic equations, especially useful after completing the square. Once an equation is expressed in the form \((x - h)^2 = k\), the square root method makes solving straightforward.

Here's how it works:

• Isolate the squared term: Ensure that \((x - h)^2\) is by itself on one side of the equation.
• Equate to a positive \(k\): Sometimes you need to add or subtract terms to achieve this.
• Take the square root: Eliminate the square by applying a square root to both sides of the equation. Remember that taking the square root generates both positive and negative solutions, symbolized by \(\pm\).
• Solve for \(x\): Simplify further to obtain the final solutions.

Using this method, you attain two potential solutions for \(x\), incorporating both possibilities when solving \((x-h)^2 = k\). In our scenario, after simplifying to \((x - \frac{5}{4})^2 = \frac{1}{16}\), taking the square root yields solutions \(x = \frac{5}{4} \pm \frac{1}{4}\). This method is powerful, serving as a bridge from completing the square to successfully solving the equation.