Problem 38 - Brief Applied Calculus

Question

Solve each equation by factoring or the Quadratic Formula, as appropriate. $$ x^{2}-3 x=54 $$

Step-by-Step Solution

Answer

The solutions are \( x = 9 \) and \( x = -6 \).

Step 1: Move All Terms to One Side

To use either factoring or the quadratic formula, we need to have the equation set to zero. We start with the equation: \[ x^2 - 3x = 54 \]Subtract 54 from both sides to get:\[ x^2 - 3x - 54 = 0 \]

Step 2: Identify Coefficients for Quadratic Formula

The general form of a quadratic equation is \( ax^2 + bx + c = 0 \). In our equation, \( a = 1 \), \( b = -3 \), and \( c = -54 \). These coefficients will be used in the quadratic formula.

Step 3: Apply the Quadratic Formula

The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the coefficients (\( a = 1 \), \( b = -3 \), \( c = -54 \)) into the formula:\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \]

Step 4: Simplify Under the Square Root

Next, calculate the expression under the square root:\[ (-3)^2 - 4 \cdot 1 \cdot (-54) = 9 + 216 = 225 \]

Step 5: Solve for x

Now substitute back into the quadratic formula:\[ x = \frac{3 \pm \sqrt{225}}{2} \]Since the square root of 225 is 15, we have:\[ x = \frac{3 \pm 15}{2} \]

Step 6: Find the Two Possible Solutions

Calculate the values for \( x \):1. \[ x = \frac{3 + 15}{2} = \frac{18}{2} = 9 \]2. \[ x = \frac{3 - 15}{2} = \frac{-12}{2} = -6 \]

Step 7: Conclude with Solution Set

The solutions to the equation \( x^2 - 3x = 54 \) are \( x = 9 \) and \( x = -6 \). This means the equation has two distinct real solutions: \( x = 9 \) and \( x = -6 \).

Key Concepts

Factoring

Factoring is a method used to solve quadratic equations by expressing the equation as a product of its factors. The equation is set to zero, then one tries to express it in the form \( (x - p)(x - q) = 0 \).Here, \( p \) and \( q \) are the roots of the equation. To solve the equation by factoring, follow these steps:

• First, ensure all terms are on one side of the equation, with zero on the other. For instance, the problem starts as \( x^2 - 3x = 54 \), and we modify it to \( x^2 - 3x - 54 = 0 \).
• Next, look for two numbers that multiply to the constant term \( c \) (-54 here) and add up to the coefficient of the \( x \) term \( b \) (-3 here).
• Once such numbers are found, rewrite the middle term (\( -3x \)) using these numbers and factor by grouping.

However, in this exercise, factoring would pose a challenge as finding such numbers is not straightforward, leading us to use the quadratic formula instead. This underscores factoring as a specialized, often trial and error-based approach, best reserved for simpler, intuitive numbers.

Quadratic Formula

The quadratic formula is a standard method used to find the solutions of quadratic equations when factoring is not viable. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This is derived from the general form of a quadratic equation: \( ax^2 + bx + c = 0 \).In our example \( x^2 - 3x - 54 = 0 \),

• we identify \( a = 1 \), \( b = -3 \), and \( c = -54 \).
• Substitute these values into the formula, and simplify the expression under the square root, known as the discriminant.
• The discriminant \( b^2 - 4ac \) determines the nature of the roots. Here, a discriminant value of 225 means two real and distinct roots.

The quadratic formula is a powerful, reliable method, often used because it guarantees a solution whether factoring is difficult or easy.

Solving Equations

Solving equations is a core skill in algebra, focusing on finding the value(s) of the unknown variable. For quadratic equations, like\( x^2 - 3x = 54 \),the process involves:

• First, rearranging all terms to one side, creating a standard form \( ax^2 + bx + c = 0 \).
• Using appropriate methods such as factoring or the quadratic formula accordingly.

Each method has its merits:- Factoring is quick for easily factorable equations,- The quadratic formula is comprehensive, providing clear solutions when the simpler approach isn't effective.Considering both approaches equips you with strategies for varied equation forms. This exercise, demonstrating how the quadratic formula efficiently solved \( x = 9 \) and \( x = -6 \), shows the importance of flexibility in choosing your method. Practice with both methods ensures readiness for any quadratic equation you might face, strengthening your problem-solving skills.