Problem 24 - Basic Technical Mathematics with Calculus

Question

In Exercises \(11-30,\) solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$2+6 v=9 v^{2}$$

Step-by-Step Solution

Answer

The solutions are \( v = \frac{1 + \sqrt{3}}{3} \) and \( v = \frac{1 - \sqrt{3}}{3} \).

Step 1: Rewrite the Equation in Standard Form

First, we need to rewrite the given equation in the standard quadratic form, which is \( ax^2 + bx + c = 0 \). The original equation is \( 9v^2 - 6v - 2 = 0 \). This is obtained by rearranging the terms from \( 2 + 6v = 9v^2 \) to \( 9v^2 - 6v - 2 = 0 \).

Step 2: Divide All Terms by Leading Coefficient

To make completing the square easier, divide every term by the leading coefficient, \( 9 \), thus transforming the equation into \( v^2 - \frac{2}{3}v - \frac{2}{9} = 0 \).

Step 3: Move the Constant to the Other Side

Isolate the constant term by moving it to the right side: \( v^2 - \frac{2}{3}v = \frac{2}{9} \). This sets us up to complete the square.

Step 4: Complete the Square

To complete the square, take half of the coefficient of \( v \), which is \( -\frac{2}{3} \), and square it: \( \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \). Add this square to both sides of the equation: \( v^2 - \frac{2}{3}v + \frac{1}{9} = \frac{2}{9} + \frac{1}{9} \).

Step 5: Write the Left Side as a Perfect Square

Now, the left side of the equation \( v^2 - \frac{2}{3}v + \frac{1}{9} \) can be written as \( \left( v - \frac{1}{3} \right)^2 \), making the equation \( \left( v - \frac{1}{3} \right)^2 = \frac{3}{9} \).

Step 6: Simplify the Equation

Simplify the right side: \( \frac{3}{9} = \frac{1}{3} \), so the equation becomes \( \left( v - \frac{1}{3} \right)^2 = \frac{1}{3} \).

Step 7: Solve for \( v \)

Take the square root of both sides: \( v - \frac{1}{3} = \pm \sqrt{\frac{1}{3}} \). Add \( \frac{1}{3} \) to both sides to solve for \( v \): \( v = \frac{1}{3} \pm \frac{\sqrt{3}}{3} \). This gives the solutions \( v = \frac{1 + \sqrt{3}}{3} \) and \( v = \frac{1 - \sqrt{3}}{3} \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra. They have the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents an unknown variable. Understanding this form is crucial for solving these types of equations. Quadratic equations can arise from a variety of real-world situations like projectile motion or area problems.

• **Standard Form**: The equation looks like \( ax^2 + bx + c = 0 \). The \( a \) term is the quadratic coefficient, representing the parabola's opening width. The \( b \) term affects the parabola's direction, and \( c \) is the y-intercept.
• **Roots of Quadratic Equations**: These are the values of \( x \) that make the equation equal to zero. They can be found using various methods including factoring, completing the square, and the quadratic formula.
• **Parabolas**: The graph of a quadratic equation is a parabola, which can open upwards or downwards depending on the sign of \( a \). Its highest or lowest point is called the vertex.

Understanding these basics provides a solid foundation for tackling quadratic equations.

Factoring Quadratics

Factoring quadratics is one of the most common methods used to solve quadratic equations. It involves rewriting the quadratic equation into a form that is easier to deal with. This method works well when the quadratic equation can be easily manipulated into products of binomials.
Here’s a simple approach to understanding this process:

• **Identify Factors**: Look for two numbers that multiply to \( ac \) (the product of the coefficient of \( x^2 \) and the constant term) and add to \( b \) (the coefficient of \( x \)).
• **Rewrite the Middle Term**: Split the \( bx \) term using the numbers found in the previous step, rewriting the quadratic as a four-term expression.
• **Factor by Grouping**: Group terms to extract common factors. This will help in reaching the factored form.
• **Solve**: Once in factored form, set each binomial to zero and solve for \( x \).

This method not only simplifies the solving process but also aids in understanding the structure of quadratic expressions.

Solving Equations

Solving equations, particularly quadratic ones, is a critical skill in algebra. Each equation presents a unique puzzle that can often be tackled with multiple strategies. One such strategy is completing the square, which provides an alternative when factoring is difficult.
In completing the square, you make a perfect square trinomial on one side of the equation to easily solve for the variable. Let's break down this method:

• **Standardize**: Start with the equation in the form \( ax^2 + bx + c = 0 \). Rearrange as necessary to isolate the quadratic terms on one side.
• **Complete the Square**: Take half of the \( b \) term, square it, and add to both sides of the equation. This forms a trinomial that can be written as a squared binomial.
• **Solve using Square Roots**: Once in a completed square form, take the square root of both sides and solve for the variable. Don’t forget to consider both the positive and negative roots.

By mastering these steps, you’ll add another powerful tool to your math toolkit for solving various equations.