Problem 66 - Algebra for College Students

Question

Find all real number solutions for each equation. $$54-6 x^{2}=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 3 \) and \( x = -3 \).

Step 1: Move Constants to Right Side

Start by isolating the quadratic term. To do this, subtract 54 from both sides of the equation: \[ 54 - 6x^2 = 0 \] becomes \[ -6x^2 = -54 \].

Step 2: Divide by Coefficient of x²

Divide both sides of the equation by -6 to solve the quadratic term. This simplifies the equation to: \[ x^2 = 9 \].

Step 3: Apply Square Root

Take the square root of both sides to solve for \( x \). Remember that this yields two solutions: \[ x = \pm \sqrt{9} \].

Step 4: Simplify Results

Calculate the square root of 9 to find \( x \). The solutions are: \[ x = 3 \] and \[ x = -3 \].

Key Concepts

Real Number Solutions

Real number solutions refer to the values of the variable that satisfy the given equation and are real numbers. In quadratic equations, solutions are the points where the graph of the equation intersects the x-axis. These solutions can be:

• Real and distinct (two different solutions)
• Real and equal (one solution repeated, also called a double root)
• No real solutions (when solutions are complex or imaginary)

For the equation \(54 - 6x^2 = 0\), we found that the real number solutions are \(x = 3\) and \(x = -3\). The equation has two distinct real number solutions, as the parabola described by this equation intersects the x-axis at these two points. Identifying and finding real number solutions is crucial because it indicates where the equation holds true under real number constraints.

Square Root Method

The square root method is a straightforward approach for solving quadratic equations of the form \(x^2 = k\), where \(k\) is a constant. This method involves the following steps:

• Isolate the squared term, \(x^2\), on one side of the equation.
• Take the square root of both sides to solve for \(x\).
• Consider both the positive and negative roots, as both are possible solutions.

In the problem \(-6x^2 = -54\), we simplified it to \(x^2 = 9\). Then, by applying the square root method, we found:

• \(x = \sqrt{9} = 3\)
• \(x = -\sqrt{9} = -3\)

This method is efficient when the equation can be easily rewritten in a form where the quadratic term is isolated. It is generally used for simpler quadratic equations that do not contain linear terms.

Solving Quadratics by Factoring

Solving quadratics by factoring involves expressing the equation in a product of two binomials set to zero and then solving each binomial equation for the variable. This method is based on the zero product property, which states if a product of two numbers is zero, at least one of the factors must be zero. Here are the steps:

• Rewrite: Ensure the equation is in the standard form \(ax^2 + bx + c = 0\).
• Factor the quadratic: Find two numbers that multiply to \(ac\) and add to \(b\).
• Set each factor to zero: Solve for the variable \(x\) by setting each binomial equal to zero.

Although the equation \(54 - 6x^2 = 0\) has already been solved using the square root method, it can also be approached by rearranging it to standard form and factoring if possible. For example, after simplifying to \(6x^2 - 54 = 0\) and factoring out the common factor, we get \(6(x^2 - 9) = 0\). Then further factoring \(x^2 - 9\) as \((x-3)(x+3)\), set each factor to zero, giving the same solutions \(x = 3\) and \(x = -3\). This makes factoring an alternative powerful tool, especially when dealing with more complex equations.