Problem 25 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$6 x^{2}=13 x-6$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3}{2}\) and \(x = \frac{2}{3}\).

Step 1: Move all terms to one side

Rearrange the equation so that it equals zero. Start by subtracting \(13x\) and subtracting \(6\) from both sides of the equation: \[6x^2 - 13x + 6 = 0\]

Step 2: Identify A, B, and C

Now that the equation is in the form \(ax^2 + bx + c = 0\), identify \(a = 6\), \(b = -13\), and \(c = 6\).

Step 3: Find two numbers that multiply to AC and add to B

You need to find two numbers that multiply to \(ac = 6 \times 6 = 36\) and add to \(b = -13\). These numbers are \(-4\) and \(-9\), because \(-4 \times -9 = 36\) and \(-4 + -9 = -13\).

Step 4: Rewrite the equation using the found numbers

Use the numbers found in Step 3 to rewrite the middle term \(-13x\): \[6x^2 - 4x - 9x + 6 = 0\]

Step 5: Factor by grouping

Group the terms into pairs and factor each pair: \[2x(3x - 2) - 3(3x - 2) = 0\] \ Notice that \((3x - 2)\) is a common factor.

Step 6: Factor out the common term

Factor out the common binomial \((3x - 2)\) from the equation: \[(2x - 3)(3x - 2) = 0\]

Step 7: Solve each factor for x

Set each factor equal to zero and solve for \(x\): 1. \(2x - 3 = 0\) leads to \(x = \frac{3}{2}\) 2. \(3x - 2 = 0\) leads to \(x = \frac{2}{3}\)

Key Concepts

Quadratic Formula

The quadratic formula is a valuable tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). Although the original exercise emphasizes factoring, understanding the quadratic formula enriches your toolkit, especially when factoring is challenging.

The formula is given by:

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

This formula helps find both potential solutions for \(x\) in a quadratic equation. These solutions occur when the expression under the square root, the discriminant \(b^2 - 4ac\), is non-negative.

When to Use the Quadratic Formula

Use this formula when:

• The quadratic equation cannot be easily factored.
• You need precise solutions, especially if solutions are irrational numbers.
• To verify factored results.

Understanding how to apply this formula will allow you to solve a broader range of quadratic equations efficiently.

Polynomial Equations

Polynomial equations may include quadratic equations, characterized by polynomials of various degrees. The quadratic equation \(6x^2 - 13x + 6 = 0\) is a specific case of a polynomial equation where the highest exponent of \(x\) is 2.

Polynomial equations can often be identified by:

• Terms of various degrees, usually ordered from highest to lowest.
• Coefficients, which could be integers, fractions, or irrational numbers.
• Constants not multiplied by the variable.

Solving Polynomial Equations

Depending on the degree and specific composition, polynomial equations can be solved by

• Factoring,
• Using algebraic identities, or
• Applying numerical methods for higher-degree polynomials.

Studying polynomial equations hone skills in pattern recognition and problem-solving, foundational in various mathematical applications.

Algebraic Solutions

Algebraic solutions involve finding the values for variables that satisfy the given equations. For quadratic equations, this often leads to specific values for \(x\) that make the equation true.

Methods for Finding Algebraic Solutions

Some algebraic strategies include:

• Factoring, where you express the equation in terms of simpler "factors" that multiply to zero.
• Completing the square to make part of the equation a perfect square trinomial.
• Using the quadratic formula, as previously introduced.

Each method has its advantages, depending on how the equation is structured.

Importance of Algebraic Solutions

Finding algebraic solutions is crucial for:

• Verifying solutions by substituting them back into the original equation.
• Understanding relationships between variables in different mathematical contexts.
• Ensuring solutions make sense within real-world applications, such as physics problems.

Grasping these techniques eases progression to more complex mathematical challenges, providing a solid foundation in algebra.