Problem 28 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by completing the square. $$\pi^{2} y^{2}+2 \pi y=3$$

Step-by-Step Solution

Answer

Solutions are \(y = \frac{1}{\pi}\) and \(y = -\frac{3}{\pi}\).

Step 1: Divide the Equation by the Leading Coefficient

The given quadratic equation is \( \pi^2 y^2 + 2\pi y = 3 \). First, divide all terms by \( \pi^2 \) to simplify the equation:\[y^2 + \frac{2}{\pi}y = \frac{3}{\pi^2}\]

Step 2: Move the Constant to the Right Side

Next, move the constant term to the right side of the equation to set up for completing the square:\[y^2 + \frac{2}{\pi}y = \frac{3}{\pi^2}\] The constant is already on the right side, so no change is needed here.

Step 3: Complete the Square

Take half of the coefficient of the \(y\) term (\(\frac{2}{\pi}\)), square it, and add it to both sides to complete the square:Half of \(\frac{2}{\pi}\) is \(\frac{1}{\pi}\), and squaring it gives \(\frac{1}{\pi^2}\). Add \(\frac{1}{\pi^2}\) to both sides:\[y^2 + \frac{2}{\pi}y + \frac{1}{\pi^2} = \frac{3}{\pi^2} + \frac{1}{\pi^2}\]This simplifies to:\[\left(y + \frac{1}{\pi}\right)^2 = \frac{4}{\pi^2}\]

Step 4: Solve for y

Take the square root of both sides of the equation. Remember to consider both the positive and negative roots:\[y + \frac{1}{\pi} = \pm \frac{2}{\pi}\]Solve for \(y\):\[y = -\frac{1}{\pi} \pm \frac{2}{\pi}\]This results in two potential solutions:\[y = \frac{1}{\pi}\quad \text{or} \quad y = -\frac{3}{\pi}\]

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation that has the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Such equations have a distinctive feature because the variable \( x \) is raised to the second power (hence the name *quadratic*, deriving from "quad" meaning square).
Quadratic equations can have:

• Two real solutions
• One real solution
• No real solutions (when they result in complex numbers)

These equations are commonly encountered in many scientific and engineering contexts, often modeling real-world phenomena. The techniques to solve them include factoring, using the quadratic formula, or, as in our exercise, completing the square.
Solving quadratics involves understanding the interactions between the terms by employing various algebraic methods to extract the values of \( x \) that satisfy the equation.

Algebra

Algebra is the branch of mathematics that deals with symbols and the rules for manipulating those symbols. It's a unifying thread of almost all mathematics and is widely used in solving various types of equations, like our quadratic equations. In algebra, symbols (often letters) represent numbers and these symbols are arranged to form expressions and equations.
When solving algebraic problems, such as quadratic equations, you often perform operations to rearrange and simplify expressions as follows:

• Adding or subtracting terms across an equation
• Multiplying or dividing terms, especially to simplify parts of the equation with coefficients
• Utilizing distributive, associative, and commutative properties to reorganize terms in more solvable forms

Algebraic problem-solving requires understanding these rules and properties to effectively manipulate equations. Completing the square, as seen in the exercise, is a useful technique to solve specific kinds of algebraic equations, allowing us to transform and simplify them to more manageable forms.

Mathematical Solution Steps

Completing the square is an algebraic process used to convert quadratic equations into a form that makes them easier to solve. Here's how it works, using the exercise as a guide:
1. **Divide by the Leading Coefficient:** Start by dividing every term by the coefficient of \( y^2 \) to simplify the equation. This step is crucial for reshaping the equation into a manageable form.2. **Rearrange Terms:** Place terms involving the variable on one side and constants on the other, prepping the expression for completing the square.3. **Add the Square:** To complete the square, take half of the coefficient of the \( y \) term, square it, and add this result to both sides. This makes the left side a perfect square trinomial, which is easier to solve:

• Half the coefficient of \( y \) is squared and added to each side.
• This creates a perfect square trinomial on the left, forming \((y + p)^2\).

4. **Solve for \( y \):** By taking the square root of both sides, remember to include both the positive and negative roots, derived from the quadratic nature of the square operation.
This meticulous step-by-step approach ensures clarity and correctness in finding solutions to the quadratic equation.