Problem 11 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by factoring. $$ 3 x^{2}-75=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 5\) and \(x = -5\).

Step 1: Write the Equation in Standard Form

The equation given is \(3x^2 - 75 = 0\). It's already in the standard form of a quadratic equation \(ax^2 + bx + c = 0\) with \(a = 3\), \(b = 0\), and \(c = -75\).

Step 2: Factor Out the Greatest Common Factor

Identify and factor out the greatest common factor (GCF) from the terms on the left side. Here, the GCF of \(3x^2\) and \(-75\) is 3. Factor 3 out of the equation: \(3(x^2 - 25) = 0\).

Step 3: Recognize and Apply the Difference of Squares Formula

Notice that \(x^2 - 25\) is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\). Here, \(x^2 - 25\) becomes \((x - 5)(x + 5)\).

Step 4: Solve the Factored Equation

Now the equation is fully factored: \(3(x - 5)(x + 5) = 0\). Set each factor equal to zero and solve for \(x\):1. \(x - 5 = 0\) \(\Rightarrow x = 5\)2. \(x + 5 = 0\) \(\Rightarrow x = -5\).

Step 5: Write the Solution

The solutions to the equation \(3x^2 - 75 = 0\) are \(x = 5\) and \(x = -5\). These are the x-values that satisfy the original quadratic equation.

Key Concepts

Factoring

Factoring is a powerful method often used to simplify and solve polynomial equations, including quadratic equations. When tackling a quadratic equation through factoring, the goal is to rewrite the equation as a product of simpler expressions. This process can help identify the solutions, or roots, of the equation.

• First, rewrite the equation if necessary so that it resembles the standard form of a quadratic: \(ax^2 + bx + c = 0\).
• Identify if there's a greatest common factor (GCF) that can be factored out from all terms.
• Once the GCF is factored out, look to factor the remaining expression, which often collapses the equation into the product of two binomials.

In our example, the quadratic equation \(3x^2 - 75 = 0\) had a GCF of 3, leading to the expression \(3(x^2 - 25)\). Factoring is a strategic step in simplifying equations—to eventually finding the roots through setting each factor to zero.

Difference of Squares

The difference of squares is a specific pattern in algebra where a squared number is subtracted by another squared number. It takes the form \(a^2 - b^2\), and it can be factored into \((a - b)(a + b)\). This technique is a key element of factoring.

• The recognition of the difference of squares allows for rapid simplification of quadratic expressions.
• Look for terms that are perfect squares, which can help in quickly applying the formula.
• This approach turns potentially complex re-arrangements into straightforward factorizations.

In the equation \(x^2 - 25\) evident in our factored quadratic, rooted squares are directly visible: \(x^2\) and \(5^2\). By identifying this pattern, we can rewrite the expression as \((x - 5)(x + 5)\). The difference of squares is a popular method because it simplifies solving and understanding quadratic equations.

Solving Quadratics

Solving quadratic equations is a fundamental skill in algebra. These equations are polynomials of degree two, and they usually take the form \(ax^2 + bx + c = 0\). Multiple methods exist for finding the values of \(x\) that satisfy the equation, amongst which factoring is particularly straightforward when applicable.

• After rewriting the equation in standard form, if factorable, express the quadratic as the multiplication of two binomials or simpler terms.
• Set each binomial expression to zero which turns the procedure into solving basic linear equations.
• Look for specific integer values that satisfy each expression, these are the solutions.

For the equation \(3(x - 5)(x + 5) = 0\), this results in the factors \(x - 5 = 0\) and \(x + 5 = 0\), with solutions \(x = 5\) and \(x = -5\). By using this method, you effectively break down the main equation into simple calculations that are solvable with basic algebra. This structured approach is vital for systematically tackling any quadratic problem.