Problem 45 - Algebra A Combined Function

Question

Solve each equation. $$ (2 x+3)\left(2 x^{2}-5 x-3\right)=0 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Zero Product Property

The equation is given as \[ (2x + 3)(2x^2 - 5x - 3) = 0. \]According to the zero product property, if the product of two expressions is zero, at least one of the expressions must be zero. Hence, we can set each factor equal to zero individually and solve for \(x\).

Step 2: Solve the First Factor

First, solve the equation \[ 2x + 3 = 0. \]Subtract 3 from both sides to get:\[ 2x = -3. \]Divide both sides by 2:\[ x = -\frac{3}{2}. \]

Step 3: Solve the Second Factor by Factorization

Next, solve the quadratic equation \[ 2x^2 - 5x - 3 = 0. \]We will factor this quadratic expression. We need two numbers whose product is \(2 \times -3 = -6\) and whose sum is \(-5\). These numbers are \(-6\) and \(1\).Rewrite the equation as:\[ 2x^2 - 6x + x - 3 = 0. \]

Step 4: Group and Factor

Group the terms:\[ (2x^2 - 6x) + (x - 3) = 0. \]Factor out the greatest common factor from each group:\[ 2x(x - 3) + 1(x - 3) = 0. \]Now factor by grouping:\[ (2x + 1)(x - 3) = 0. \]

Step 5: Solve for x from the Factors

Apply the zero product property again to solve:1. \[ 2x + 1 = 0 \] Subtract 1 from both sides: \[ 2x = -1 \] Divide by 2: \[ x = -\frac{1}{2}. \]2. \[ x - 3 = 0 \] Add 3 to both sides: \[ x = 3. \]

Step 6: Compile All Solutions

The solutions to the equation \[ (2x + 3)(2x^2 - 5x - 3) = 0 \]are given by:\[ x = -\frac{3}{2}, \; x = -\frac{1}{2}, \; x = 3. \]

Key Concepts

Factoring Quadratics

To solve quadratic equations effectively, one important skill is factoring them. Quadratic equations are any polynomial equation of the form \( ax^2 + bx + c = 0 \). The goal of factoring is to express this equation in the product of two binomials or simpler expressions. This way, it becomes much easier to solve.Factoring essentially breaks down the quadratic into manageable parts. Consider the expression \( 2x^2 - 5x - 3 \):

• First, determine the product of the leading coefficient (2) and the constant term (-3). This gives us -6.
• Next, we look for two numbers that multiply to -6 and add up to the middle term, -5. These numbers are -6 and 1.

With these numbers identified, rewrite the quadratic as \( 2x^2 - 6x + x - 3 \). From here, you can group terms and factor each group to arrive at the factored form: \( (2x + 1)(x - 3) = 0 \). This step is foundational, setting up the equation so you can use further problem-solving strategies.

Solving Equations

Once a quadratic equation has been factored, the next step is to solve for the variable. This process involves finding the values of \( x \) that satisfy the equation.Take our example equation, \( (2x + 3)(2x^2 - 5x - 3) = 0 \). After factoring the quadratic part as \( (2x + 1)(x - 3) \), we set each factor equal to zero:

• \( 2x + 3 = 0 \)
• \( 2x + 1 = 0 \)
• \( x - 3 = 0 \)

This property allows us to solve three simpler linear equations. Solving these successively will yield each potential value of \( x \). As a result, solving equations by factoring facilitates determining the solutions straightforwardly.

Zero Product Rule

Crucial to solving factored equations, the Zero Product Rule is straightforward: if the product of two numbers is zero, then at least one of the numbers must be zero. This logical principle underpins why we can separate a factorized equation into simpler linear equations.For instance, in \( (2x + 3)(2x^2 - 5x - 3) = 0 \), apply the Zero Product Rule:

• Either \( 2x + 3 = 0 \) or \( 2x^2 - 5x - 3 = 0 \) must be true.

Once you solve the factorized quadratic to \( (2x + 1)(x - 3) = 0 \), each of these factors can individually equal zero:

• Set \( 2x + 1 = 0 \) and solve for \( x \).
• Set \( x - 3 = 0 \) and solve for \( x \).

This rule simplifies the process, ensuring that no solutions are overlooked and providing the foundation upon which factoring operates.

Quadratic Equation Solutions

The final step in working with quadratic equations is finding their solutions, the values of \( x \) that satisfy the original equation. These solutions are called the roots of the equation.From the factorization and solving steps, each value of \( x \) is considered a solution. For our equation, after factoring and applying the zero product rule:

• From \( 2x + 3 = 0 \), we find \( x = -\frac{3}{2} \).
• From \( 2x + 1 = 0 \), we discover \( x = -\frac{1}{2} \).
• From \( x - 3 = 0 \), we uncover \( x = 3 \).

Each of these values, \( x = -\frac{3}{2}, x = -\frac{1}{2}, x = 3 \), is a solution to the quadratic equation. Understanding the mechanics behind quadratic solutions transforms solving them from a complex task into a series of manageable, logical steps.