Problem 52 - Algebra 2

Question

Solve each equation by completing the square. \(x^{2}-\frac{2}{3} x-\frac{26}{9}=0\)

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{3} + \sqrt{3}\) and \(x = \frac{1}{3} - \sqrt{3}\).

Step 1: Identify Coefficients

For the quadratic equation in the form of \(ax^2 + bx + c = 0\), identify \(a = 1\), \(b = -\frac{2}{3}\), and \(c = -\frac{26}{9}\).

Step 2: Move Constant to the Other Side

Rewrite the equation by moving the constant term to the other side: \(x^2 - \frac{2}{3}x = \frac{26}{9}\).

Step 3: Complete the Square

To complete the square, take half of \(b\), which is \(-\frac{2}{3}\), divide it by 2 to get \(-\frac{1}{3}\), and finally square it to obtain \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\). Add \(\frac{1}{9}\) to both sides of the equation: \(x^2 - \frac{2}{3}x + \frac{1}{9} = \frac{26}{9} + \frac{1}{9}\).

Step 4: Simplify Both Sides

Simplify the right side: \(x^2 - \frac{2}{3}x + \frac{1}{9} = 3\); this is achieved by adding \(\frac{26}{9}\) and \(\frac{1}{9}\) to get \(\frac{27}{9} = 3\).

Step 5: Write as a Perfect Square

Factor the left side as a perfect square: \((x - \frac{1}{3})^2 = 3\).

Step 6: Solve for x

Take the square root of both sides: \(x - \frac{1}{3} = \pm \sqrt{3}\).

Step 7: Solve for Each Case

Solve for \(x\) by isolating it in each case: \(x = \frac{1}{3} + \sqrt{3}\) or \(x = \frac{1}{3} - \sqrt{3}\).

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra that involves expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The 'quadratic' part of the name comes from the Latin word for 'square' since the variable is squared. This squaring leads to a curve called a parabola when graphed on the Cartesian plane. Understanding quadratic equations is key as they can model various real-world phenomena, like projectile motion and area problems.

In our exercise, the quadratic equation is \(x^2 - \frac{2}{3}x - \frac{26}{9} = 0\). Here, we identify:\

• \(a = 1\)
• \(b = -\frac{2}{3}\)
• \(c = -\frac{26}{9}\)

Recognizing these coefficients sets the stage for techniques like completing the square, which is a common method to find the values of \(x\) that make the equation true.

Perfect Square Trinomial

A perfect square trinomial is the result of squaring a binomial and is expressed in the form \((x + d)^2\) where \(d\) is a number. Expanding this will give us \(x^2 + 2dx + d^2\), a trinomial consisting of a quadratic, linear, and constant term.

The idea of completing the square revolves around transforming a quadratic expression into a perfect square trinomial. Why? Because a perfect square is easy to solve as it is just one term squared.

In the exercise, after moving the constant term to the other side and arranging the equation to look like \(x^2 - \frac{2}{3}x = \frac{26}{9}\), the task is to add a number to both sides to make the left side a perfect square trinomial. By taking half of the coefficient of \(x\) (\(-\frac{2}{3}\)), dividing by 2, and then squaring it, we find that number is \(\frac{1}{9}\). Adding \(\frac{1}{9}\) to both sides results in \(x^2 - \frac{2}{3}x + \frac{1}{9}\), which factors neatly into \((x - \frac{1}{3})^2\). This simplifies solving immensely!

Solving Equations

Solving equations, especially quadratic ones, can be approached through different methods. Completing the square is particularly useful for understanding the geometric interpretation and solving equations when other methods may be cumbersome.

Once we rewrite the equation as a perfect square trinomial, \((x - \frac{1}{3})^2 = 3\), taking the square root of both sides simplifies the equation dramatically. This step helps isolate \(x\) by addressing the square directly:\

• \(x - \frac{1}{3} = \sqrt{3}\)
• \(x - \frac{1}{3} = -\sqrt{3}\)

Solving these resulting linear equations, we find the solutions to the original quadratic equation to be \(x = \frac{1}{3} + \sqrt{3}\) and \(x = \frac{1}{3} - \sqrt{3}\). This step not only refines your algebra skills but also increases your problem-solving toolkit for broader mathematical applications.