Problem 26 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by completing the square. $$2+6 v=9 v^{2}$$

Step-by-Step Solution

Answer

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

Step 1: Rearrange the Equation

To begin solving the quadratic equation by completing the square, we need to have the quadratic in the standard form: \[ ax^2 + bx + c = 0 \] The given equation is: \[ 2 + 6v = 9v^2 \] Subtract \(6v\) and \(2\) from both sides to get it into standard form: \[ 9v^2 - 6v - 2 = 0 \] This rearranges the equation to a standard quadratic form.

Step 2: Divide by the Leading Coefficient

The leading coefficient \( a \) in our equation is 9. To simplify completing the square, we divide every term by 9: \[ v^2 - \frac{2}{3}v - \frac{2}{9} = 0 \] This gives a simpler equation to work with: \[ v^2 - \frac{2}{3}v = \frac{2}{9} \]

Step 3: Complete the Square

Next, we need to complete the square for the left-hand side of the equation: 1. Take the coefficient of \( v \), which is \(-\frac{2}{3}\), divide by 2: \[ \frac{-\frac{2}{3}}{2} = -\frac{1}{3} \] 2. Square this result: \[ \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \]3. Add this square inside the equation to balance both sides: \[ v^2 - \frac{2}{3}v + \frac{1}{9} = \frac{2}{9} + \frac{1}{9} \]This simplifies to: \[ (v - \frac{1}{3})^2 = \frac{3}{9} = \frac{1}{3} \]

Step 4: Solve for v

Now solve for \( v \) by taking the square root of both sides. Remember to consider both the positive and negative roots: \[ v - \frac{1}{3} = \pm \frac{1}{\sqrt{3}} \] Add \( \frac{1}{3} \) to both sides to solve for \( v \): \[ v = \frac{1}{3} \pm \frac{1}{\sqrt{3}} \] Rationalize the denominator if needed: \[ v = \frac{1}{3} + \frac{\sqrt{3}}{3}\quad \text{or} \quad v = \frac{1}{3} - \frac{\sqrt{3}}{3} \]

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They appear in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. These equations are fundamental in algebra because they represent parabolic graphs and have interesting properties in their solutions.

• The solutions to quadratic equations can be found using several methods, such as factoring, using the quadratic formula, or completing the square.
• Quadratics can have two real roots, one real root, or two complex roots, depending on the discriminant \( b^2 - 4ac \).

It's crucial to understand how quadratic equations differ from other polynomial equations due to their degree's specific properties.

Standard Form of Quadratic Equation

The standard form of a quadratic equation is essential for applying various algebraic techniques. It is expressed as \( ax^2 + bx + c = 0 \). When a quadratic equation like \( 2 + 6v = 9v^2 \) is presented, we must first arrange it into the standard form.

• Bring all terms to one side to set the equation equal to zero.
• Reorder the terms to align with \( ax^2 + bx + c = 0 \).

For example, our equation becomes \( 9v^2 - 6v - 2 = 0 \) after rearranging. This step is vital as it allows for easier manipulation and application of techniques such as completing the square.

Solving Quadratic Equations

To solve quadratic equations, multiple methods can be employed, each with its advantages. Completing the square is one effective technique that transforms the equation into a perfect square trinomial.

• This method involves expressing the quadratic in the form \((v - p)^2 = q\) through careful manipulation and balancing.
• We can then take square roots of both sides to solve for the variable.

For instance, from the equation \( v^2 - \frac{2}{3}v = \frac{2}{9} \), completing the square transforms it into \((v - \frac{1}{3})^2 = \frac{1}{3} \). Thus, solving becomes straightforward after this transformation.

Algebraic Techniques

Algebraic techniques such as completing the square require a solid grasp of algebra fundamentals. This technique is useful when the equation is hard to factor or when precision is required. Here's how it works effectively:

• Identify and manipulate the quadratic and linear terms to complete the square.
• Add and subtract the same value to keep the equation balanced.

In our example, this process involves dividing \(-\frac{2}{3}\) by two and squaring it, resulting in a term \(\frac{1}{9}\), which balances the equation. Thus, the left side becomes a perfect square that can be easily solved. Understanding these detailed steps strengthens your ability to deal with quadratic equations smoothly.