Problem 43 - Algebra 2

Question

Solve each equation by using the Square Root Property. \(x^{2}+1.4 x+0.49=0.81\)

Step-by-Step Solution

Answer

The solutions are \(x = 0.2\) and \(x = -1.6\).

Step 1: Move Constant to One Side

Start by moving the constant term on the right side of the equation, so that the equation is set to zero. Subtract 0.81 from both sides to obtain \(x^2 + 1.4x + 0.49 - 0.81 = 0\), which simplifies to \(x^2 + 1.4x - 0.32 = 0\).

Step 2: Complete the Square

Since the equation is not initially a perfect square trinomial, adjust the equation to make it a perfect square on the left. Because we have \(x^2 + 1.4x\), take half of 1.4, square it, and add it to both sides. Half of 1.4 is 0.7, and \(0.7^2 = 0.49\). Thus, the equation becomes \(x^2 + 1.4x + 0.49 = 0.81\).

Step 3: Apply Square Root Property

Remove the complete square by applying the square root property. Since \((x + 0.7)^2 = 0.81\), take the square root of both sides, resulting in \(x + 0.7 = \pm 0.9\).

Step 4: Solve for x

From \(x + 0.7 = 0.9\), subtract 0.7 from both sides, yielding \(x = 0.9 - 0.7 = 0.2\). Or from \(x + 0.7 = -0.9\), subtract 0.7, giving \(x = -0.9 - 0.7 = -1.6\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that always have the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). The solution of a quadratic equation typically results in two values of \(x\), known as the roots of the equation.

These equations are fundamental in algebra and appear frequently in various mathematical contexts. They can represent anything from the path of a projectile to finding the maximum area given a perimeter constraint. Understanding quadratic equations is key to solving problems efficiently in both mathematics and real-life applications.

• **Nature of Roots**: The roots of a quadratic equation can be real or complex depending on the value of the discriminant \(b^2 - 4ac\).
• **Real and Equal**: If \(b^2 - 4ac = 0\), the roots are real and equal.
• **Real and Distinct**: If \(b^2 - 4ac > 0\), the roots are real and distinct.
• **Complex**: If \(b^2 - 4ac < 0\), the roots are complex.

Recognizing the form and behavior of quadratic equations prepares you to apply different solving techniques like factoring, the quadratic formula, completing the square, or using the Square Root Property.

Completing the Square

Completing the square is a technique you can use to solve quadratic equations by transforming the equation into a perfect square trinomial. This form makes it easier to solve by hand, especially when the equation doesn't factor easily.

To complete the square, consider the equation \(x^2 + bx + c = 0\). The goal is to transform the left side into a square, such as \((x + p)^2\). Follow these steps:

• Move the constant \(c\) to the other side of the equation.
• Take half of the coefficient \(b\), square it, and add the result to both sides of the equation.
• Rewrite the left side as a square: \((x + p)^2 = q\).

This process is applied in our example, where we took half of 1.4, resulting in 0.7, and squared it to obtain 0.49. Adding this to both sides created a perfect square trinomial. By rewriting \(x^2 + 1.4x + 0.49\) as \((x + 0.7)^2\), we simplified the equation and made it straightforward to apply the square root property.

Solving Equations

Once a quadratic equation is expressed in the form of a perfect square trinomial using the "completing the square" method, solving it becomes easier with the application of the Square Root Property. This property states that if \((x + p)^2 = q\), then \(x + p\) is equal to the positive or negative square root of \(q\).

In our specific equation, after completing the square, we arrived at \((x + 0.7)^2 = 0.81\). By applying the Square Root Property, we took the square root of both sides to find two possible values:

• \(x + 0.7 = 0.9\)
• \(x + 0.7 = -0.9\)

By solving these simple linear equations:

• Subtracting 0.7 from each side of the equation \(x + 0.7 = 0.9\) gives us \(x = 0.2\)
• Subtracting 0.7 from each side of \(x + 0.7 = -0.9\) results in \(x = -1.6\)

Thus, the solutions to the original quadratic equation are \(x = 0.2\) and \(x = -1.6\). Solving equations with these techniques provides a systematic method to find roots, making complex problems manageable.