Problem 39 - Introductory Algebra for College Students

Question

Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$4 x^{2}-25=0$$

Step-by-Step Solution

Answer

The solutions to the given quadratic equation \(4 x^{2}-25=0\) are \(x = 5/2\) and \(x = -5/2\).

Step 1: Write Down The Quadratic Equation

The given quadratic equation is \(4 x^{2}-25=0\). Our goal is to factorize this equation in order to solve for \(x\).

Step 2: Factorize the Quadratic Equation

Notice that \(4x^{2}-25\) is a difference of squares, which factors as \((a-b)(a+b)\). In this case, \(a\) would be \(2x\) and \(b\) would be \(5\). So, \(4x^{2}-25\) can be factorized as \((2x-5)(2x+5)\). So, the equation becomes \((2x-5)(2x+5)=0\).

Step 3: Solve for Each Factor

Setting each factor to zero and solving for \(x\) gives the solutions. When we set \(2x-5 = 0\), we solve for \(x\) to get \(x = 5/2\). When we set \(2x+5 = 0\), we solve for \(x\) to get \(x = -5/2\).

Step 4: Check the Solution

Substitute \(x = 5/2\) and \(x = -5/2\) into the original equation to check the solution. The original equation becomes \(4*((5/2)^2)-25=0\) and \(4*((-5/2)^2)-25=0\), both of which are indeed equal to 0. Therefore, \(x = 5/2\) and \(x = -5/2\) are indeed the solutions to the equation.

Key Concepts

Difference of Squares

The difference of squares is a specific pattern that occurs in quadratic equations. It refers to expressions like \(a^2 - b^2\). This type of expression can be factored into \((a-b)(a+b)\), making it a useful tool in solving certain algebra problems.

For example, in the quadratic equation \(4x^2 - 25 = 0\), each term is a perfect square: \(4x^2\) and \(25\). The square root of \(4x^2\) is \(2x\) and the square root of \(25\) is \(5\). So, \(4x^2 - 25\) becomes \((2x-5)(2x+5)\) when factored using the difference of squares method.

This technique of recognizing the difference of squares can simplify solving quadratic equations and is widely used in algebra.

• Identify perfect squares on each side of the minus sign.
• Apply the formula: \(a^2 - b^2 = (a-b)(a+b)\).
• Factor and solve each resulting equation separately.

Solving Quadratic Equations

Solving quadratic equations can be done through various methods, such as factoring, using the quadratic formula, completing the square, or graphing.

In the case of quadratic equations that are a difference of squares, factoring is often the simplest and most direct method. Once the quadratic expression is factored into two binomials like \((2x-5)(2x+5)\), the next step is to solve for \(x\) using the zero product property.

The zero product property states that if the product of two numbers equals zero, then at least one of the numbers must be zero. Thus, you can set each factor equal to zero to solve for the variable:

• Set \(2x - 5 = 0\) and solve for \(x\) to get \(x = 5/2\).
• Set \(2x + 5 = 0\) and solve for \(x\) to get \(x = -5/2\).

This approach ensures that you find all possible solutions to the quadratic equation.

Factoring Techniques

Factoring is a crucial technique in algebra, particularly for solving equations. While there are several factoring techniques, one of the quickest ways to solve certain quadratics is by recognizing special patterns like the difference of squares or perfect square trinomials.

For example:

• Difference of Squares: Identifying and using this common pattern simplifies many problems, especially when dealing with expressions like \(4x^2 - 25\), which factor into \((2x-5)(2x+5)\).
• Factoring by grouping: This method is useful for quadratics that are not simple differences of squares or when equations become more complex.
• Trial and Error: This involves testing various factor combinations to determine the correct binomial factors that solve the equation.

Grasping these techniques increases problem-solving efficiency and capability. Practicing different methods boosts understanding and confidence, allowing for effective solutions to more complex algebraic equations.