Problem 31 - Elementary Algebra

Question

For the following problems, solve the equations using the quadratic formula. $$ (a+4)(a-5)=2 $$

Step-by-Step Solution

Answer

Answer: The two possible values for 'a' are \(a_1 = \frac{1 + \sqrt{89}}{2}\) and \(a_2 = \frac{1 - \sqrt{89}}{2}\).

Step 1: Expanding the equation to standard form

To expand the equation, we will first multiply out \((a+4)(a-5)\) and subtract 2 from both sides to make the equation equal to 0. $$ (a+4)(a-5) = 2 \\ a^2 - 5a + 4a -20 = 2 \\ a^2 - a - 22 = 0 $$ Now our quadratic equation is in standard form, \(a^2 - a - 22 = 0\). The coefficients are \(a=1, b=-1,\) and \(c=-22\).

Step 2: Applying the quadratic formula

Next, we'll plug the coefficients into the quadratic formula to find the values of 'a'. $$ a = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-22)}}{2(1)} \\ a = \frac{1 \pm \sqrt{1+88}}{2} \\ a = \frac{1 \pm \sqrt{89}}{2} $$ The two possible values for 'a' are: $$ a_1 = \frac{1 + \sqrt{89}}{2}, \quad a_2 = \frac{1 - \sqrt{89}}{2} $$ Thus, the solutions to the given quadratic equation are: $$ a_1 = \frac{1 + \sqrt{89}}{2}, \quad \textrm{and} \quad a_2 = \frac{1 - \sqrt{89}}{2}. $$

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two. This means its highest power of the variable, often denoted as \( x \) or some other letter, is squared. These equations are standard in algebra and appear frequently in various scientific contexts. Quadratic equations often take one of the following forms: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
These equations are particularly interesting because they have a parabolic graph, resulting in a U-shape. Depending on the values of \( a \), \( b \), and \( c \), the vertex of this parabola can be moved and its direction can be altered (upward or downward).
Solving a quadratic equation can be done using several methods, including factoring, completing the square, or, as in this exercise, the quadratic formula. The quadratic formula provides a reliable method to find the solutions or roots of any quadratic equation.

Standard Form

The standard form of a quadratic equation is crucial for solving it using the quadratic formula. This form is typically written as \( ax^2 + bx + c = 0 \). Here, the equation is set to zero after any necessary algebraic manipulation, like expanding brackets or moving terms across the equation.
In our exercise, we begin with the equation \((a+4)(a-5) = 2\). To convert this into standard form, we distribute the terms:

• Expand \((a+4)(a-5)\) to get \(a^2 - 5a + 4a - 20\).
• Simplify by combining like terms to get \(a^2 - a - 20\).
• Subtract 2 from both sides to achieve \(a^2 - a - 22 = 0\).

Now, the equation is in the standard form, making it ready for application of the quadratic formula.

Coefficients

In a quadratic equation written in standard form, \( ax^2 + bx + c = 0 \), the coefficients are \( a \), \( b \), and \( c \). These constants play a significant role in determining the nature of the roots and the shape of the graphed equation.
In our converted equation \( a^2 - a - 22 = 0 \):

• \( a \) is the coefficient of the squared term, here \( 1 \).
• \( b \) is the coefficient of the first-degree term \( a \), here \(-1\).
• \( c \) is the constant term, here \(-22\).

When applying the quadratic formula, these coefficients are used as follows:\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Understanding the role of each coefficient helps in determining how the equation behaves and makes solving simpler. For example, the term \( b^2 - 4ac \) is called the discriminant, which helps in predicting whether the roots are real or complex.