Problem 56 - Elementary Algebra

Question

\(c^{2}-12 c-28=0\)

Step-by-Step Solution

Answer

The solutions are \(c = 14\) and \(c = -2\).

Step 1: Identify the coefficients

Identify the coefficients from the quadratic equation in the form of \(ax^2 + bx + c = 0\). For the equation \(c^2 - 12c - 28 = 0\), \(a = 1\), \(b = -12\), and \(c = -28\).

Step 2: Use the quadratic formula

The quadratic formula is given by \(c = \frac{-b \,\pm\, \sqrt{b^2 - 4ac}}{2a} \). Use the identified coefficients to plug into the formula: \(c = \frac{-(-12) \,\pm\, \sqrt{(-12)^2 - 4(1)(-28)}}{2(1)}\).

Step 3: Simplify inside the square root

First, calculate \((-12)^2\) to get 144, then calculate \(4(1)(-28)\) to get -112. Simplify inside the square root: \(c = \frac{12 \,\pm\, \sqrt{144 + 112}}{2} \).

Step 4: Calculate the discriminant

Add the values inside the square root: \(144 + 112 = 256\), then take the square root of 256 to get 16. Now the equation is simplified to: \(c = \frac{12 \,\pm\, 16}{2}\).

Step 5: Find the two solutions

Using the plus sign: \(c = \frac{12 + 16}{2} = \frac{28}{2} = 14\). Using the minus sign: \(c = \frac{12 - 16}{2} = \frac{-4}{2} = -2\). So, the two solutions are \(c = 14\) and \(c = -2\).

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a second-degree polynomial equation in a single variable, usually denoted as x. The general form is \(ax^2 + bx + c = 0\). The highest exponent of x is 2, which makes it quadratic.
A quadratic equation can have one, two, or no real solutions, depending on certain conditions. Solving these equations often involves methods like factoring, completing the square, or using the quadratic formula.

Exploring Coefficients

In the quadratic equation \(ax^2 + bx + c = 0\), the coefficients are the numbers that multiply the variables. These are typically known as a, b, and c.
Here’s a breakdown:

• \(a\)
  is the coefficient of \(x^2\), the quadratic term.
• \(b\)
  is the coefficient of \(x\), the linear term.
• \(c\)
  is the constant term, having no variable attached.

For example, in \(c^2 - 12c - 28 = 0\), the equation can be rewritten as \(1c^2 - 12c - 28 = 0\), giving you:

• \(a = 1\)
• \(b = -12\)
• \(c = -28\)

Understanding the Discriminant

The discriminant is a key component in the quadratic formula, represented by \(b^2 - 4ac\). It helps determine the nature and number of the solutions (roots) of a quadratic equation.
Depending on the value of the discriminant, you can have:

• Discriminant > 0: Two distinct real solutions (roots).
• Discriminant = 0: One real solution; a repeated root.
• Discriminant < 0: No real solutions; two complex solutions.

For our given equation, \(c^2 - 12c - 28 = 0\), the discriminant calculation would be: \((-12)^2 - 4(1)(-28) = 144 + 112 = 256\). Since 256 is greater than zero, this equation has two distinct real solutions.

Solving Using Square Root

The quadratic formula \(c = \frac{-b \,\pm\, \sqrt{b^2 - 4ac}}{2a}\) relies heavily on the square root operation to simplify the result.
Taking the square root is part of simplifying the discriminant and affects the final solutions.
Here’s how you handle square roots in our context:
After calculating the discriminant from \(b^2 - 4ac\) and getting 256, you need to find its square root. The \(\sqrt{256} = 16\).
This step is crucial as it leads to the core solving formula:
\(c = \frac{12 \,\pm\, 16}{2}\). Thus, the square root simplifies the problem down to easily manageable operations to find the final solutions, resulting in two answers:\(c = 14\) and\(c = -2\).