Problem 30 - College Algebra and Calculus: An Applied Approach

Question

Use the Quadratic Formula to solve the quadratic equation. $$ (x+6)^{2}=-2 x $$

Step-by-Step Solution

Answer

The solutions of the equation \((x+6)^{2}=-2 x\) are \(x = -7+sqrt(13)\) and \(x = -7-sqrt(13)\)

Step 1: Expand and Simplify

Expand the square and simplify the equation to get it in the standard quadratic form. \((x+6)^{2}=-2 x\rightarrow x^{2} + 12x + 36 = -2x\). Transfer all terms to one side to get the equation to the standard form: \(x^{2} + 12x + 2x + 36 = 0\), which simplifies to \(x^{2} + 14x + 36 = 0\)

Step 2: Identify Parameters and Define Quadratic Formula

Identify the parameters a, b, and c in the simplified equation, where \(a=1\), \(b=14\), and \(c=36\). The Quadratic Formula is \(x = [-b\pm sqrt(b^{2} - 4ac)] / (2a)\)

Step 3: Apply Quadratic Formula

Substitute the values of a, b, and c into the formula: \(x = [-14 \pm sqrt((14)^{2} - 4*1*36)] / (2*1) = [-14\pm sqrt(196 - 144)] / 2 = [-14\pm sqrt(52)] / 2\). Simplify this to get the solutions: \(x = -7\pm sqrt(13)\)

Step 4: Final Answer

Thus the solutions are \(x = -7+sqrt(13)\) and \(x = -7-sqrt(13)\)

Key Concepts

Solving Quadratic Equations

The process of finding the values of the variable that satisfy the equation is known as solving quadratic equations. These equations are of the form \( ax^2 + bx + c = 0 \) where \( a \) is not equal to zero. One common method to find the solutions is by using the Quadratic Formula. This approach is particularly useful for equations that do not factor easily.

For the given exercise, we begin by expanding and simplifying the given equation \( (x+6)^2 = -2x \) to achieve the standard quadratic form. After simplification, we apply the Quadratic Formula with the identified coefficients \( a, b, \) and \( c \). It is vital to ensure that no arithmetic errors occur during this step, as precision is key to obtaining the correct solutions. Substituting the values gives us two potential solutions, as the formula accommodates both the addition and subtraction of the square root term.

Standard Quadratic Form

The standard quadratic form of an equation is expressed as \( ax^2 + bx + c = 0 \) where \( a \) is the coefficient of \( x^2 \) and is non-zero, \( b \) is the coefficient of \( x \) and \( c \) is the constant term. To solve a quadratic equation using the Quadratic Formula or by factoring, the equation must first be in this standard form.

In the given problem, we saw the initial equation \( (x+6)^2 = -2x \) transformed into the standard form \( x^2 + 14x + 36 = 0 \) by expanding the square and moving all terms to one side. This is a crucial step in the solving process, as it lays the groundwork for all subsequent solution methods.

Discriminant of a Quadratic

The discriminant of a quadratic equation provides valuable information about the nature and number of solutions without actually solving the equation. It is represented by the expression \( b^2 - 4ac \) found under the square root in the Quadratic Formula. Depending on the value of the discriminant, we can predict the following:

• If \( b^2 - 4ac > 0 \) there are two real and distinct solutions.
• If \( b^2 - 4ac = 0 \) there is one real, repeated solution.
• If \( b^2 - 4ac < 0 \) there are two complex solutions.

For our equation \( x^2 + 14x + 36 = 0 \) the discriminant is \( 14^2 - 4(1)(36) \) which simplifies to \( 52 \) indicating there are two real and distinct solutions which we find using the Quadratic Formula. Understanding the discriminant is a crucial part of solving quadratics as it provides a shortcut to determining the solution set.