Problem 33 - Elementary Algebra

Question

Solve by completing the square. $$x 2-0.6 x-0.27=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0.9\) and \(x = -0.3\).

Step 1: Rewrite in Standard Form

The given equation is \(x^2 - 0.6x - 0.27 = 0\). It's already in the form \(ax^2 + bx + c = 0\). Identify \(a = 1\), \(b = -0.6\), and \(c = -0.27\).

Step 2: Move the Constant to the Other Side

Rearrange the equation to isolate the constant term: \(x^2 - 0.6x = 0.27\).

Step 3: Find the Value to Complete the Square

Take half of the coefficient of \(x\) and square it. Half of \(-0.6\) is \(-0.3\), and squaring gives \((-0.3)^2 = 0.09\).

Step 4: Add and Subtract to Complete the Square

Add \(0.09\) to both sides: \(x^2 - 0.6x + 0.09 = 0.27 + 0.09\).

Step 5: Simplify the Equation

Simplify the right side: \(x^2 - 0.6x + 0.09 = 0.36\).

Step 6: Rewrite as a Perfect Square

The left side can be rewritten as \((x - 0.3)^2 = 0.36\).

Step 7: Solve for x

Take the square root of both sides: \(x - 0.3 = \pm \sqrt{0.36}\). Simplifying gives \(x - 0.3 = \pm 0.6\).

Step 8: Find the Possible Values of x

Solving \(x - 0.3 = 0.6\) yields \(x = 0.9\). Solving \(x - 0.3 = -0.6\) yields \(x = -0.3\).

Key Concepts

Quadratic Equations

To solve problems involving quadratic equations, you must first understand what they are and how they function. A quadratic equation is any equation that can be rearranged into the standard form \(ax^2 + bx + c = 0\). The quadratic term is \(x^2\), which means the highest power of the variable \(x\) is 2. This form is important because it helps us identify each coefficient: \(a\), \(b\), and \(c\).

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term

Recognizing these components is the first step in manipulating quadratic equations for solutions. In our example, we identified \(a = 1\), \(b = -0.6\), and \(c = -0.27\). This understanding will guide us through solving using various methods, one of which is completing the square.

Perfect Square

When dealing with quadratic equations, completing the square involves creating a perfect square trinomial from the equation. A perfect square trinomial is one which can be expressed as the square of a binomial. For example, \((x - 0.3)^2\) is a perfect square trinomial in the given exercise.Creating a perfect square involves adding and subtracting the same value to the equation. This new addition transforms the quadratic into something easier to solve. In our example, we started with \(x^2 - 0.6x\) and completed the square by determining that the value to add is \(0.09\), derived from half of \(-0.6\) squared. This allowed us to rewrite the left side as \((x - 0.3)^2\), making it a perfect square. This is a critical step because it prepares the equation for straightforward solution techniques, like taking the square root of both sides.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to make them easier to solve. Completing the square is one type of algebraic manipulation for solving quadratic equations. Key steps involve:

• Identifying coefficients \(b\)
• Calculating half of \(b\) and squaring it
• Adding and subtracting that result to form a perfect square trinomial

In our exercise, we rearranged the equation to \(x^2 - 0.6x = 0.27\). Then, we completed the square to make \((x - 0.3)^2 = 0.36\). This process transforms the equation, making it much easier and more intuitive to solve.Solve the Equation:
The final step in algebraic manipulation here is solving for \(x\) by taking the square root of both sides of the equation. Knowing that \((x - 0.3)^2 = 0.36\), we got:
\(x - 0.3 = \pm 0.6\). Therefore:

• When \(x - 0.3 = 0.6\), thus \(x = 0.9\)
• When \(x - 0.3 = -0.6\), thus \(x = -0.3\)

Practicing these steps in rearranging and completing squares will greatly enhance your ability to solve quadratic equations efficiently.