Problem 38 - Algebra 1

Question

Solve the equation by completing the square. $$x^{2}-\frac{2}{3} x-3=0$$

Step-by-Step Solution

Answer

The solution to the equation \(x^{2}-\frac{2}{3} x-3=0\) is \(x= -\frac{1}{3} \pm\sqrt{-3+\frac{1}{9}}\). Note that the roots are imaginary if the term under the square root is negative.

Step 1: Identify \(a\), \(b\), and \(c\)

In the equation \(x^{2}-\frac{2}{3} x-3=0\), \(a=1\), \(b=-\frac{2}{3}\), and \(c=-3\).

Step 2: Complete the square

The term to complete the square is \((b/2)^{2} = ((-\frac{2}{3})/2)^{2} = (-\frac{1}{3})^{2} = \frac{1}{9}\). Add this term to both sides of the equation: \(x^{2}-\frac{2}{3} x+ \frac{1}{9}=-3 + \frac{1}{9}\).

Step 3: Form a perfect square trinomial

The left hand side is now a perfect square trinomial, and can be written as \((x-b/2)^{2}\), where \(b=-\frac{2}{3}\). This gives us: \((x+\frac{1}{3})^{2} = -3+\frac{1}{9}\).

Step 4: Solve for x

Solve for x by taking the square root of both sides remember to consider both positive and negative roots. This gives: \(x+\frac{1}{3}= \pm\sqrt{-3+\frac{1}{9}}\). Now isolate x to get: \(x= -\frac{1}{3} \pm\sqrt{-3+\frac{1}{9}}\).

Key Concepts

Quadratic Equations

Quadratic equations form one of the foundational concepts in mathematics, especially within algebra. They get their name because their highest degree term is "quadratus," which means square in Latin. These equations are usually written in the form \(ax^2 + bx + c = 0\). Here, \(x\) is the variable, and \(a\), \(b\), and \(c\) are constants with \(a eq 0\). In the provided exercise, the quadratic equation is \(x^2 - \frac{2}{3}x - 3 = 0\).

• To solve quadratic equations, you can use methods like factoring, using the quadratic formula, or completing the square.
• Completing the square helps transform the equation into a form, making it easier to find the variable's value.

You'll encounter quadratic equations in various scenarios, from calculating areas and projectile motions to solving numerous real-world problems.By understanding their format and finding their roots or solutions, you unlock the basics of essential algebraic concepts.

Perfect Square Trinomial

Perfect square trinomials are special algebraic expressions that simplify the process of solving quadratic equations through completing the square. A trinomial is an algebraic expression composed of three terms. A perfect square trinomial, specifically, is one that can be expressed as the square of a binomial.Let's take an example from our exercise: \(x^2 - \frac{2}{3}x + \frac{1}{9}\).

• Here, this trinomial can be rewritten as \((x - \frac{1}{3})^2\).
• This is accomplished by adjusting the equation such that it perfectly fits the form \(a^2 - 2ab + b^2 = (a - b)^2\).

The process involves determining the missing term that perfectly completes the square.Adding \((\frac{b}{2})^2\) ensures that the left side of the equation becomes a perfect square trinomial.This makes solving for the variable much simpler as it can be expressed as the square of a binomial, leading to finding the roots of the quadratic equation efficiently.

Solving Equations

Once the quadratic equation has been transformed into a perfect square trinomial, the next step is solving for \(x\). By rewriting the equation in this way, we're in a position to take the square root of both sides to isolate \(x\). Here's how it works using our example: 1. We have \((x - \frac{1}{3})^2 = -3 + \frac{1}{9}\).2. Simplify the right side: This becomes \((x - \frac{1}{3})^2 = -\frac{26}{9}\).3. Take the square root of both sides:

• This will give \(x - \frac{1}{3} = \pm \sqrt{-\frac{26}{9}}\).
• The \(\pm\) symbol is used because squaring can introduce two possible values (positive or negative).

4. Finally, isolate \(x\), leading to \(x = -\frac{1}{3} \pm i\sqrt{\frac{26}{9}}\).Completing the square can sometimes introduce imaginary numbers if the number inside the square root is negative. Nonetheless, this method provides an organized approach to find solutions for quadratic equations effectively.