Problem 56 - Elementary Algebra

Question

Solve by extracting the roots and then solve by using the quadratic formula. Check answer. $$ (x+3) 2+9=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -3 + 3i \) and \( x = -3 - 3i \).

Step 1: Simplify the Equation

Start by expanding the equation \( (x+3)^2 + 9 = 0 \). By expanding we have \((x+3)^2 = x^2 + 6x + 9\). Therefore the equation becomes \( x^2 + 6x + 9 + 9 = 0 \) which simplifies to \( x^2 + 6x + 18 = 0 \).

Step 2: Extracting Roots Method

Attempt to solve the equation \( x^2 + 6x + 18 = 0 \) by extracting roots. First, move 18 to the right side to get \( x^2 + 6x = -18 \). Then complete the square by adding \( (6/2)^2 = 9 \) to both sides to get\( x^2 + 6x + 9 = -18 + 9 \) which is \( (x+3)^2 = -9 \). Taking the square root of both sides gives \( x+3 = \pm \sqrt{-9} \) or \( x = -3 \pm 3i \).

Step 3: Quadratic Formula Method

Use the quadratic formula to solve \( x^2 + 6x + 18 = 0 \) where \( a=1, b=6, \text{ and } c=18 \). The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). First, calculate the discriminant: \( b^2 - 4ac = 6^2 - 4(1)(18) = 36 - 72 = -36 \). Thus, the roots are \( x = \frac{-6 \pm \sqrt{-36}}{2} \) which simplifies to \( x = \frac{-6 \pm 6i}{2} \), giving \( x = -3 \pm 3i \).

Step 4: Verify the Solutions

Both methods produced the solutions \( x = -3 + 3i \) and \( x = -3 - 3i \). Therefore, these are the solutions to the original equation.

Key Concepts

Complex Numbers

When solving quadratic equations, there are times when you might end up with a negative number inside the square root. This is where complex numbers come into play. A complex number is of the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit. Imaginary numbers arise because the square root of a negative number is not defined in the set of real numbers. Hence, the imaginary unit \( i \) is defined as \( i^2 = -1 \).

This concept becomes essential when working with quadratic equations that result in a negative discriminant. The discriminant \( b^2 - 4ac \) dictates the nature of the roots of the equation. A negative value implies that the roots will be complex numbers. For instance, in our equation, the discriminant was \(-36\), leading to roots \( x = -3 \pm 3i \). Understanding complex numbers allows one to handle such roots confidently and accurately.

Completing the Square

Completing the square is a method used to solve quadratic equations by transforming them into a perfect square trinomial. This approach makes it easier to solve the equation, especially when other methods like factoring are difficult or impossible. Here's a simple breakdown:

• Begin with a standard quadratic equation \( ax^2 + bx + c = 0 \).
• Ensure that the coefficient of \( x^2 \) is 1. If not, divide the entire equation by \( a \).
• Next, take the coefficient of \( x \), divide it by 2, and square it. Add and subtract this number inside the equation.

Returning to our problem, after simplifying \( x^2 + 6x + 18 = 0 \) to \( x^2 + 6x = -18 \), we complete the square by adding \( 9 \) (since \( (6/2)^2 = 9 \)). This transforms it to \((x + 3)^2 = -9 \). Taking the square root yields \( x + 3 = \pm \sqrt{-9} \). This illustrates how completing the square easily leads us to the solution.

Quadratic Formula

The quadratic formula is a powerful tool that solves any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is expressed as:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula is derived from completing the square and provides direct roots without further manipulation. Let's break down the steps:

• Check the coefficients \( a, b, \) and \( c \).
• Calculate the discriminant \( b^2 - 4ac \). This determines the nature of the roots (real or complex).
• Substitute the values into the formula.

In our exercise, the values are \( a = 1, b = 6, \) and \( c = 18 \). The discriminant was \(-36\), confirming complex roots. Plugging into the formula results in \( x = -3 \pm 3i \), matching with our other solution method. The quadratic formula ensures consistent results across different quadratic situations.