Problem 8 - College Algebra

Question

Solve the quadratic equation by factoring. $$ 2 x^{2}+9 x-5=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = -5\) and \(x = \frac{1}{2}\).

Step 1: Identify the quadratic equation

The given quadratic equation is \(2x^2 + 9x - 5 = 0\). Our task is to solve it by factoring.

Step 2: Multiply a and c (AC Method)

First, identify coefficients: \(a = 2\), \(b = 9\), and \(c = -5\). Multiply \(a\) and \(c\): \(2 \times (-5) = -10\).

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

Find two numbers that multiply to \(-10\) and add to \(9\). These numbers are \(10\) and \(-1\) because \(10 \times (-1) = -10\) and \(10 + (-1) = 9\).

Step 4: Rewrite the middle term

Rewrite the equation \(2x^2 + 9x - 5\) as \(2x^2 + 10x - x - 5\).

Step 5: Factor by grouping

Group the terms: \((2x^2 + 10x) + (-x - 5)\). Factor out the common factors: \(2x(x + 5) - 1(x + 5)\).

Step 6: Factor out the common binomial factor

Since \((x + 5)\) is a common factor, factor it out: \((x + 5)(2x - 1) = 0\).

Step 7: Solve each factor set to zero

Set each factor to zero: \(x + 5 = 0\) and \(2x - 1 = 0\).Solve for \(x\):1. \(x + 5 = 0\) gives \(x = -5\).2. \(2x - 1 = 0\) gives \(2x = 1\), so \(x = \frac{1}{2}\).

Key Concepts

Factoring

When we talk about factoring in relation to quadratic equations, we're addressing the process of breaking down an expression into products of simpler expressions. In simpler terms, it's like looking for pairs of numbers or terms that multiply together to give our original equation. It's an essential step in solving quadratic equations.To factor a quadratic equation, such as \( ax^2 + bx + c = 0 \), we want to express it as \((px + q)(rx + s) = 0\). The equation then becomes the product of two binomials. Once in this form, we can easily solve for the variable \( x \) by using the zero product property, which states that if a product of two factors equals zero, at least one of the factors must be zero.Factoring may not always be straightforward. Fortunately, methods such as the AC Method and Factor by Grouping can make it easier to find those pairs. Understanding these methods helps simplify quadratic expressions, making it easier and faster to find solutions.

AC Method

The AC Method is a special trick that helps in factoring when the leading coefficient \( a \) is not 1. It's particularly useful in quadratic equations of the form \( ax^2 + bx + c = 0 \), where you might find direct factoring difficult.Here's the step-by-step process:

• First, multiply \( a \) (the coefficient of \( x^2 \)) with \( c \) (the constant term). This product, called the \( AC \) product, guides us in finding the right numbers.
• Second, find two numbers that multiply to give the \( AC \) product and add up to \( b \) (the middle coefficient).
• These two numbers are key because they help us split the \( bx \) term effectively, enabling us to proceed to the next step: factoring by grouping.

The AC Method is a strategic helper that breaks down more complex quadratics into manageable segments, paving the way for successful factoring.

Factoring by grouping

Factoring by grouping is a method that uses the two numbers found using the AC Method to simplify the problem into groups that can be factored easily. This method simplifies the equation so that it's easier to manage.For example, with the equation \( 2x^2 + 9x - 5 \), after finding the numbers 10 and -1 through the AC Method:

• Rewrite the equation by splitting the middle term into two terms. In this case, the equation becomes \( 2x^2 + 10x - x - 5 \).
• Group the terms to prepare for individual factoring. Here, this results in two groups: \((2x^2 + 10x)\) and \((-x - 5)\).
• Next, factor out the greatest common factor from each group. In \( 2x^2 + 10x \), you can factor out \( 2x \), and in \(-x - 5\), you factor out \(-1\).

Once you have factored each group individually, you will see a common binomial factor (in this example, \( x + 5 \)). This commonality allows you to factor further into \((x + 5)(2x - 1)\), setting up the final step for solving the equation.

Zero product property

The zero product property is a simple yet powerful tool that plays a central role in solving quadratic equations. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. This logical property allows for straightforward solutions to factored quadratic equations.Once a quadratic equation is fully factored, like in our example \((x + 5)(2x - 1) = 0\), you can apply the zero product property by setting each factor equal to zero:

• First, set \( x + 5 = 0 \). Solve this to find \( x = -5 \).
• Next, set \( 2x - 1 = 0 \). Solve this to find \( x = \frac{1}{2} \).

These solutions, \( x = -5 \) and \( x = \frac{1}{2} \), are the solutions to the quadratic equation. Using the zero product property turns a potentially complex equation into a simple one by leveraging the factors identified through the preceding steps.