Problem 13 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation by factoring. $$-3 x^{2}+12=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation are \(x = 2\) and \(x = -2\).

Step 1: Rewrite Equation

First, rewrite the equation as \(-3 x^{2} + 12 = 0\) to \(3x^{2} - 12 = 0\). Here, both terms were divided by -1 to simplify the equation.

Step 2: Factor the equation

Factor the equation into two terms. \(3x^{2} - 12 = 0\) is equivalent to \(3(x^{2} - 4) = 0\). Now factor the difference of squares inside the parenthesis to get \((x - 2)(x + 2)\). Multiply it back to the outside term 3 to get the final factored form as \(3(x - 2)(x + 2) = 0\)

Step 3: Solve for x

Now, set each factor equal to zero, and then solve for x. This will give the solutions to the equation. For the factor \(x - 2\), the equation is \(x - 2 = 0\), and solving for x gives \(x = 2\). For the factor \(x + 2\), the equation is \(x + 2 = 0\), and solving for x gives \(x = -2\). ${\rm So} x=2$ or $x=-2$ is the solution.

Key Concepts

Factoring

Factoring is like finding a pair of glasses that helps you see numbers and expressions more clearly by breaking them down into simpler parts. When we factor a quadratic equation like \(-3x^2 + 12 = 0\), we look for a way to express it as a product of simpler expressions.

• First, simplify the equation by eliminating the negative sign. This step involves changing \(-3x^2 + 12 = 0\) to \(3x^2 - 12 = 0\) by dividing everything by -1.
• Next, observe if there's a common factor in the simplified equation terms. Here, both terms 3 and 12 are divisible by 3.
• Factor out the common factor. So the equation becomes \(3(x^2 - 4) = 0\).

The question now transforms into an easier problem about finding two numbers (\(x - 2\) and \(x + 2\)) that when multiplied together, along with 3, will give zero. Factoring reduces a complex equation into bite-sized pieces making it easier to handle and solve.

Difference of Squares

The difference of squares is a special type of factoring where an expression is made of two perfect squares separated by a minus sign. It follows the pattern \(a^2 - b^2 = (a-b)(a+b)\).
In the context of our quadratic equation \(3(x^2 - 4) = 0\), the expression \(x^2 - 4\) is a difference of squares:

• Recognize \(x^2 - 4\) as a difference of squares since \(x^2\) and \(4\) (which is \(2^2\)) are perfect squares.
• Apply the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \), where \(a = x\) and \(b = 2\).

This means \((x^2 - 4) = (x - 2)(x + 2)\).
The difference of squares allows us to quickly and easily break down equations that might seem complex at first glance.Exploit this property to simplify and solve expressions effectively by recognizing this pattern.

Solving Equations

After factoring a quadratic equation successfully, solving it becomes straightforward. This involves finding the values of \( x \) that satisfy the equation.
Here's how it's done:

• Set each factor equal to zero. From \(3(x - 2)(x + 2) = 0\), you focus on the expressions inside the parenthesis: \(x - 2\) and \(x + 2\).
• For \(x - 2\), set it equal to zero: \(x - 2 = 0\). Solving for \(x\) gives \(x = 2\).
• For \(x + 2\), set it equal to zero: \(x + 2 = 0\). Solving for \(x\) gives \(x = -2\).

Both values of \( x \) satisfy the equation, so \(x = 2\) and \(x = -2\) are the solutions.Solving quadratic equations via factoring involves splitting the problem into smaller pieces, solving each part, and reuniting their solutions.