Problem 20 - Beginning and Intermediate Algebra

Question

Solve using the zero-factor property. $$ 8 x^{2}+18 x+9=0 $$

Step-by-Step Solution

Answer

x = - \frac{3}{4} or x = - \frac{3}{2}.

Step 1 - Identify the given quadratic equation

The given quadratic equation is \[ 8x^{2} + 18x + 9 = 0 \]

Step 2 - Factorize the quadratic equation

To factorize the given equation, look for two binomials whose product gives the quadratic equation. Notice that \[ 8x^2 + 18x + 9 \] can be rewritten as \[ (4x + 3)(2x + 3) \].

Step 3 - Apply the zero-factor property

According to the zero-factor property, if \[ (4x + 3)(2x + 3) = 0 \], then at least one of the factors must be zero.

Step 4 - Solve for x from each factor

Set each factor to zero and solve for x:For \[ 4x + 3 = 0 \], \[ 4x = -3 \] \[ x = - \frac{3}{4} \]For \[ 2x + 3 = 0 \], \[ 2x = -3 \] \[ x = - \frac{3}{2} \]

Key Concepts

Factoring Quadratic Equations

Factoring quadratic equations is a crucial skill in algebra. It involves rewriting a quadratic equation in a product form. This product form usually consists of two binomials that when multiplied together, yield the original quadratic equation. For example, consider the quadratic equation \(8x^{2} + 18x + 9 = 0\). To factor this, you need to find two binomials such that their product gives the quadratic equation. In our case, \(8x^{2} + 18x + 9\) can be factored into \( (4x + 3)(2x + 3) \). This means that these two binomials multiply together to give the original quadratic equation. Understanding factoring helps simplify complex expressions and is foundational for solving quadratic equations.

Solving Quadratic Equations

Solving quadratic equations is all about finding the values of \x\ that make the equation true. Once you have factored the equation, use the zero-factor property to find these values. In our example, after factoring \(8x^2 + 18x + 9\) into \( (4x + 3)(2x + 3) \), we set each binomial to zero. This gives us two separate equations to solve:
\(4x + 3 = 0 \) and \(2x + 3 = 0\).
By solving these equations, we find the values of \x\ that satisfy the original quadratic equation. Solving \(4x + 3 = 0\) gives \( x = -\frac{3}{4}\), and solving \(2x + 3 = 0\) yields \( x = -\frac{3}{2} \). These values are our solutions to the quadratic equation.

Algebraic Solutions

Algebraic solutions involve using algebraic methods to solve equations. In the case of a quadratic equation, the primary methods include factoring, completing the square, and using the quadratic formula. For this problem, we used factoring along with the zero-factor property to find the solutions. The zero-factor property states that if the product of two expressions is zero, then at least one of the expressions must be zero. This is why we set each binomial factor to zero and solved for \( x \). Understanding and applying these algebraic methods is essential because it allows you to break down complex problems into simpler, solvable parts. With practice, these methods become powerful tools in your mathematical toolkit.