Problem 16 - Basic Technical Mathematics with Calculus

Question

solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$3 x^{2}=3-4 x$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{-2 + \sqrt{13}}{3} \) and \( x = \frac{-2 - \sqrt{13}}{3} \).

Step 1: Rearrange the Equation

First, we need to rearrange the given equation into the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). The given equation is \( 3x^2 = 3 - 4x \). Let's move all terms to one side to get: \( 3x^2 + 4x - 3 = 0 \).

Step 2: Identify Coefficients

Identify the coefficients from the standard form of the quadratic equation \( 3x^2 + 4x - 3 = 0 \). Here, \( a = 3 \), \( b = 4 \), and \( c = -3 \).

Step 3: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 3 \), \( b = 4 \), and \( c = -3 \) into this formula.

Step 4: Solve the Discriminant

Calculate the discriminant using the formula \( b^2 - 4ac \). Here, we have \( 4^2 - 4 \times 3 \times (-3) = 16 + 36 = 52 \).

Step 5: Compute the Roots

Now substitute the discriminant and other values into the quadratic formula: \[ x = \frac{-4 \pm \sqrt{52}}{6} \]. Simplify \( \sqrt{52} \) to \( 2\sqrt{13} \), resulting in two solutions: \[ x = \frac{-4 + 2\sqrt{13}}{6} \] and \[ x = \frac{-4 - 2\sqrt{13}}{6} \].

Step 6: Simplify the Expression

Further simplify the expressions for \( x \): \[ x = \frac{-2 + \sqrt{13}}{3} \] and \[ x = \frac{-2 - \sqrt{13}}{3} \].

Key Concepts

Quadratic Equations

Quadratic equations are a central concept in algebra. These are polynomial equations of degree two, meaning they have a variable raised to the power of two. Typically, they are written in the standard form:

• \( ax^2 + bx + c = 0 \)
• Here, \( a \), \( b \), and \( c \) are constants.

The coefficient \( a \) must not be zero for it to qualify as a quadratic equation. The solutions to a quadratic equation can be found by various methods such as factoring, completing the square, or using the quadratic formula. Each method has its own use depending on the nature of the equation.
For example, the equation \( 3x^2 = 3 - 4x \) can be rearranged into standard form as:

• \( 3x^2 + 4x - 3 = 0 \)

This step is crucial as it prepares the equation for further analysis using these solution techniques.

Discriminant Calculation

The discriminant is a key part of understanding the nature of the solutions to a quadratic equation. It is found within the quadratic formula itself:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Here, \( b^2 - 4ac \) is the discriminant.

Calculating the discriminant helps us determine how many and what type of solutions exist.

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution.
• If \( b^2 - 4ac < 0 \), there are no real solutions (the solutions are complex).

In our example with \( 3x^2 + 4x - 3 = 0 \), we find:

• \( b^2 - 4ac = 16 + 36 = 52 \)

This positive discriminant indicates that there are two real solutions.

Solving Quadratics

Solving quadratic equations using the quadratic formula is a reliable method that works for any quadratic equation. The formula is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This method requires you to replace \( a \), \( b \), and \( c \) with their values from the standard form of the equation. For our example:

• \( a = 3 \), \( b = 4 \), \( c = -3 \)
• Substitute these into the formula:
  • \( x = \frac{-4 \pm \sqrt{52}}{6} \)
• Simplifying further gives:
  • \( x = \frac{-2 + \sqrt{13}}{3} \) and \( x = \frac{-2 - \sqrt{13}}{3} \)

This shows two solutions, illustrating the power and utility of the quadratic formula even when other methods like factoring might not be applicable.

Mathematics Education

Understanding quadratic equations is a fundamental part of mathematics education. It helps build a foundation for more advanced areas of study, such as calculus and linear algebra.
Educators often emphasize the different methods for solving quadratics to ensure that students have a versatile set of tools.

• Factoring: Involves expressing the equation as a product of its factors.
• Completing the Square: Rewrites the equation in a form that makes finding the roots more straightforward.
• Quadratic Formula: A universal method applicable to all quadratic equations.

Teaching these concepts encourages logical thinking and problem-solving. By going through the steps of using the quadratic formula, students learn to identify, substitute, and simplify mathematical expressions. This process is integral in developing an appreciation for mathematics and enhancing critical thinking skills.