Problem 56 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation using any method. Find only real solutions. $$x^{2}-9=0$$

Step-by-Step Solution

Answer

The solution to the equation is \(x=3\) and \(x=-3\).

Step 1: Arrange the expression

We start by arranging the given quadratic equation \(x^{2}-9=0\) to \(x^{2}=9\).

Step 2: Taking square root

After this, we take a square root of both sides, giving us \(x=\sqrt{9}\) and \(x=-\sqrt{9}\).

Step 3: Calculate square root and find x

By calculating the square roots, we find the solution is \(x=3\) and \(x=-3\).

Key Concepts

Real Solutions

When solving quadratic equations, finding the real solutions refers to identifying roots of the equation that are real numbers. Real solutions are important as they represent real-world situations and can be graphed on the coordinate plane. Quadratic equations can have either:

• Two distinct real solutions
• One real solution (a repeated root)
• No real solution if the roots are complex

In the example equation, \(x^2 - 9 = 0\), we are asked to find only the real solutions. This means we are interested in the values of \(x\) where the squared term equals a positive number (in this case, 9), leading to two distinct real solutions at \(x = 3\) and \(x = -3\).
Understanding whether solutions are real, repeated, or complex involves looking at how the quadratic can be factored or solved, typically by evaluating the expression under the square root sign (b² - 4ac, known as the discriminant) in the more general quadratic formula.

Square Root Method

The square root method is a straightforward way to solve quadratic equations that have a simpler structure, like \(x^2 = n\). This method works perfectly when the quadratic equation can be reduced to a square term on one side and a constant on the other.
Here's how it works:

• First, rearrange the equation to make sure the \(x^2\) term is isolated. In our exercise, this gives us the form \(x^2 = 9\).
• Next, apply the square root to both sides of the equation. Remember, taking the square root of both sides means we consider both the positive and negative roots. Hence, \(x = \sqrt{9}\) or \(x = -\sqrt{9}\).
• Finally, calculate the square roots to determine the values of \(x\). Here, it results in \(x = 3\) and \(x = -3\).

This method is efficient for quadratics without a linear \(x\) term, quickly delivering real solutions when applicable.

Quadratic Equations

Quadratic equations are fundamental algebraic structures that involve a variable raised to the second power. They take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations are pivotal in various areas of mathematics and science.
The solutions to quadratic equations represent the points where the curve, typically a parabola, intersects the x-axis. There are several methods to solve them:

• Factoring: Works well when the quadratic easily breaks down into a product of binomial expressions.
• Completing the square: Converts the quadratic into a perfect square trinomial.
• Quadratic formula: A universal method that uses the formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) to find solutions.
• Graphing: For a visual approach to approximate roots.

The exercise \(x^2 - 9 = 0\) is an example of a simple quadratic equation with no linear term, making it easily solvable by the square root method. Real solutions for such equations are straightforward to identify and calculate.