Question

Solve the equation by completing the square. \(6 r^2+6 r+12=0\)

Short answer

The solutions to the equation are \(r = -1/2 \pm (1/2) * \sqrt{7} i\)

Step-by-step solution

Simplify the Equation

First, it's best to simplify the equation. We can do this by dividing each term by the coefficient of the \( r^2 \) term. So, divide each term in the equation by 6, thus forming the equation: \(r^2 + r + 2 = 0 \)

Move the Independent Term

Next, move the independent term (the term that does not contain the variable 'r') to the other side of the equation by subtraction. So, we have: \(r^2 + r = -2\)

Complete the Square

Now, we complete the square. To do this, we add the square of half of the coefficient of the 'r' term to each side. The coefficient of the 'r' term is 1, and half of 1 squared is 1/4. So we add 1/4 to each side of the equation: \(r^2 + r + 1/4 = -2 + 1/4\)

Simplify the Negative Side

We can simplify the right side of the equation by converting the decimal 1/4 into a fraction with the same denominator as 2 (8/4), and then adding the fractions: -2 is equivalent to -8/4 and -8/4 + 1/4 = -7/4. So the equation becomes: \(r^2 + r + 1/4 = -7/4\)

Root the Equation

We can now rewrite the left side of the equation as a perfect square and take the square root of both sides. The left side can be written as \((r + 1/2)^2\), the right side will be \(\pm\sqrt{-7/4}\). Therefore, we have: \((r + 1/2) = \pm \sqrt{-7/4}\)

Solve for 'r'

At this point we can solve for 'r'. Since we are dealing with negative square roots, the solutions will be imaginary. So, we write the square root of -7/4 as \( \pm( 1/2 ) * \sqrt{7}\) because the square root of -1 is \(i\). The solutions then become: \(r = -1/2 \pm (1/2) * \sqrt{7} i\)

Key concepts

Algebraic Equations

Algebraic equations are mathematical statements composed of polynomials that set expressions equal to a value, most commonly zero. They include variables (unknown numbers you want to find), coefficients (fixed numbers), and operations such as addition, subtraction, and multiplication.

• In our exercise, the algebraic equation given is: \(6r^2 + 6r + 12 = 0\).
• This equation is quadratic because it involves the variable \(r\) raised to the power of two.
• Solving such equations often involves strategies like factoring, using the quadratic formula, or completing the square as we'll do here.

Each algebraic equation can have different levels of complexity based on the number of variables and the degree of each term. Understanding and manipulating these elements are key to solving them.

Complex Solutions

Complex solutions arise when you solve equations that involve taking the square root of negative numbers. In essence, these solutions embrace both real and imaginary components. To complete an equation involving complex solutions, knowing how to work with these components is crucial.

• In our solution, completing the square led us to a scenario where the square root of a negative number was necessary, thus indicating complex solutions.
• The solutions to the quadratic are expressed in terms of real and imaginary parts: \(r = -1/2 \pm (1/2) \sqrt{7} i\).

This resultant form shows how combining real numbers with imaginary units (indicated by 'i') transforms a calculation that might seem impossible into something solvable and meaningful.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, usually taking the form \(ax^2 + bx + c = 0\). These are prominent in algebra due to their wide applications in various fields such as physics, engineering, and finance.

• The standard approach for solving includes methods like factoring, using the quadratic formula, or completing the square.
• In this exercise, the equation was simplified to \(r^2 + r + 2 = 0\) before applying the completing the square technique.
• Quadratics can have two, one, or no real solutions, with complex solutions arising when the discriminant (the part under the square root in the quadratic formula \(b^2 - 4ac\)) is negative.

Such equations are a foundation in mathematics due to their straightforward structure and significant implications when modeling real-world phenomena.

Imaginary Numbers

Imaginary numbers provide a way to extend our understanding of the number system to include square roots of negative numbers. The concept of imaginary numbers is crucial to solving equations that don't have real solutions.

• In our quadratic equation, when reaching \((r + 1/2)^2 = -7/4\), it became necessary to consider the square root of -1, which is defined as the imaginary unit \(i\).
• This transformed our solutions into terms that included \(i\), indicating they are not purely real but instead complex.
• Imaginary numbers, when combined with real numbers, form complex numbers, which are often expressed in the form \(a + bi\).

Mastering the use of imaginary numbers allows mathematicians and engineers to approach problems that go beyond the limits of real number solutions, offering insights and resolutions in various advanced fields.