Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by completing the square. Show each step. $$ 2+z=6 z^{2} $$

Step-by-Step Solution

Answer

The solutions are \(z = \frac{2}{3}\) and \(z = -\frac{1}{2}\).

Step 1: Rewrite the Equation

First, rewrite the given quadratic equation in standard form. The original equation given is \(2 + z = 6z^2\). Start by subtracting \(6z^2\) and \(z\) from both sides to bring all terms to one side of the equation. This gives:\[6z^2 - z - 2 = 0\]

Step 2: Divide by the Leading Coefficient

Divide every term in the quadratic equation by the leading coefficient of the squared term, which is 6, to simplify the equation. Doing this, we have:\[z^2 - \frac{1}{6}z - \frac{1}{3} = 0\]

Step 3: Isolate the Constant Term

To complete the square, move the constant term to the other side of the equation. This gives:\[z^2 - \frac{1}{6}z = \frac{1}{3}\]

Step 4: Complete the Square

Find the coefficient of the linear term, which is \(-\frac{1}{6}\), and divide it by 2, then square it. \[-\frac{1}{6} \div 2 = -\frac{1}{12}, \left(-\frac{1}{12}\right)^2 = \frac{1}{144}\]Add this square to both sides to form a perfect square trinomial:\[z^2 - \frac{1}{6}z + \frac{1}{144} = \frac{1}{3} + \frac{1}{144}\]

Step 5: Simplify and Factor the Equation

Simplify the right side by finding a common denominator:\[\frac{1}{3} = \frac{48}{144}\] so, \[\frac{48}{144} + \frac{1}{144} = \frac{49}{144}\]Now, the equation becomes:\[\left(z - \frac{1}{12}\right)^2 = \frac{49}{144}\]

Step 6: Solve for z

Take the square root of both sides of the equation to solve for \(z\):\[z - \frac{1}{12} = \pm \frac{7}{12}\]Add \(\frac{1}{12}\) to both sides to find the two solutions for \(z\):\[z = \frac{1}{12} + \frac{7}{12} = \frac{8}{12} = \frac{2}{3}\]\[z = \frac{1}{12} - \frac{7}{12} = -\frac{6}{12} = -\frac{1}{2}\]

Key Concepts

Completing the Square

Completing the square is a technique used to solve quadratic equations. This method involves creating a perfect square trinomial from the original quadratic equation, which makes it easier to solve. The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] To complete the square, follow these steps:

• Ensure the coefficient of the squared term, \(a\), is 1. If not, divide the whole equation by \(a\).
• Move the constant term, \(c\), to the other side of the equation.
• Take the linear coefficient, \(b\), divide it by 2, and then square this result. This value is added to both sides of the equation to maintain balance.

This process turns the expression into a perfect square of a binomial. For example, in the equation: \[ z^2 - \frac{1}{6}z + \frac{1}{144} = \frac{49}{144} \]the left side forms a square: \[ \left(z - \frac{1}{12}\right)^2 = \frac{49}{144} \] This allows us to easily solve for \(z\) by taking the square root of both sides.

Polynomial Equations

Polynomial equations are mathematical expressions that involve variables raised to various powers. The coefficient terms are usually constants. The most common type of polynomial is the quadratic equation, such as: \[ 6z^2 - z - 2 = 0 \]This equation is quadratic because it features the variable \(z\) raised to the second power, or squared.Polynomials come with various degrees:

• A linear polynomial, when the highest power is one (e.g., \(y = 3x + 2\)).
• A quadratic polynomial, when the degree is two (as in our case \(z^2\)).
• A cubic polynomial, when the degree is three, and so on.

Quadratics are characterized by their degree (2), and they often model many real-life scenarios, from physics trajectories to financial calculations. Polynomial equations of higher degrees tend to be more complex to solve, requiring specific methods for each degree.

Solving Equations

Solving equations involves finding the values that satisfy the equality. For quadratic equations, several methods can be used, such as factoring, using the quadratic formula, and completing the square.For the equation: \[ 6z^2 - z - 2 = 0 \] The solution involves isolating \(z\). In the given exercise, completing the square is used, which transforms the equation into a more manageable form:\[ \left(z - \frac{1}{12}\right)^2 = \frac{49}{144} \]The next step is to solve for \(z\) by calculating the square root of both sides. By doing this, you'll need to adjust for both the positive and negative square roots, as squaring either returns the same positive value.Once the perfect square trinomial is formed and square-rooted, solve for \(z\) to find the possible solutions. For instance, this results in:

• \(z = \frac{2}{3}\)
• \(z = -\frac{1}{2}\)

These solutions are verified by substituting back into the original equation to ensure they satisfy the equality. Solving equations is about consistently applying algebraic methods to simplify and determine the values of the unknown variables.