Problem 49 - Precalculus Mathematics for Calculus

Question

Solve the equation by factoring. $$3 x^{2}+5 x=2$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{1}{3} \) and \( x = -2 \).

Step 1: Rearrange the Equation

First, rearrange the equation so that one side equals zero. Start with the original equation:\[ 3x^2 + 5x = 2 \]Subtract 2 from both sides of the equation:\[ 3x^2 + 5x - 2 = 0 \]

Step 2: Identify a, b, and c in the Quadratic

The equation \( 3x^2 + 5x - 2 = 0 \) is in the standard quadratic form \( ax^2 + bx + c = 0 \), where:- \( a = 3 \)- \( b = 5 \)- \( c = -2 \)

Step 3: Factor the Quadratic

We need to factor the quadratic expression \( 3x^2 + 5x - 2 \). We will find two numbers that multiply to \( a \cdot c = 3 \times (-2) = -6 \) and add to \( b = 5 \).The numbers that meet these criteria are 6 and -1, because \( 6 \times (-1) = -6 \) and \( 6 + (-1) = 5 \).

Step 4: Write the Expression with Two Middle Terms

Use the numbers found in Step 3 to rewrite the middle term (5x) so:\[ 3x^2 + 6x - x - 2 = 0 \]

Step 5: Factor by Grouping

First, group the terms:\[ (3x^2 + 6x) + (-x - 2) = 0 \]Factor out the greatest common factor from each group:\[ 3x(x + 2) - 1(x + 2) = 0 \]

Step 6: Factor the Expression Further

Notice now we have a common binomial factor \((x + 2)\):\[ (3x - 1)(x + 2) = 0 \]

Step 7: Solve for x

Set each factor equal to zero: 1) \( 3x - 1 = 0 \)2) \( x + 2 = 0 \)Solve the first equation:\[ 3x = 1 \]\[ x = \frac{1}{3} \]Solve the second equation:\[ x = -2 \]

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). This technique involves expressing the quadratic equation as a product of two binomials. The goal is to transform the equation into a form that is easier to solve by setting each binomial equal to zero.To factor a quadratic equation:

• Ensure the equation is in standard form by moving all terms to one side so that the equation equals zero.
• Identify coefficients \( a \), \( b \), and \( c \) from the equation \( ax^2 + bx + c = 0 \).
• Find two numbers that multiply to \( a \times c \) and add up to \( b \).
• Rewrite the middle term using the two numbers found, effectively splitting it into two separate components.
• Proceed to factor the equation further, often using other techniques like grouping, to find the solution.

Factoring quadratics can simplify your solution process significantly, especially when the numbers are modest and manageable. This method is often the preferred choice for introductory algebra courses.

Quadratic Formula

The quadratic formula is a universal solver for any quadratic equation. It is particularly useful when factoring is not straightforward. The formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]allows us to find the roots, or solutions, of a quadratic equation.Using the quadratic formula involves:

• Ensuring the equation is arranged in the standard quadratic form \( ax^2 + bx + c = 0 \).
• Substituting the values of \( a \), \( b \), and \( c \) into the formula.
• Calculating the discriminant \( b^2 - 4ac \) as it dictates the nature of the roots (real and distinct, real and equal, or complex).
• Simplifying the remaining expression to find the possible values of \( x \).

While the quadratic formula is powerful and straightforward, it is computationally heavier than factoring in straightforward cases. Therefore, it is typically used when factoring is challenging or impossible.

Factoring by Grouping

Factoring by grouping is a technique often used when a quadratic equation can be written in a format that makes it possible to group terms to reveal common factors. It’s an indispensable tool, especially for quadratics that don’t fit simple factorization paths.Here's how you factor by grouping:

• With the equation in standard form, split the middle term into two terms that add up to the original middle coefficient and whose coefficients multiply to \( a \times c \).
• Group the four terms into two pairs, each capable of simplifying with a common factor.
• Factor out these common factors, which should leave you with a common binomial factor between the grouped terms.
• The final factored expression will be the product of a binomial and a remaining term.

This method is especially beneficial when dealing with quadratic expressions that arise from more complex polynomial division or synthetic division scenarios, offering a structured approach to reach solutions efficiently.