Problem 36 - Algebra A Combined Function

Question

Solve each equation by completing the square. $$ y^{2}+6 y=-8 $$

Step-by-Step Solution

Answer

The solutions are \( y = -2 \) and \( y = -4 \).

Step 1: Move constant to the other side

Start by rearranging the equation to have the variable terms on one side and the constant on the other. This means changing the given equation \( y^2 + 6y = -8 \) to \( y^2 + 6y + 8 = 0 \), which is a conceptual mistake. The correct format should keep the variable terms together: \( y^2 + 6y = -8 \). While the rearranging wasn't necessary here due to the variable setup, in reaching the form \( y^2 + 6y = -8 \), we should have aimed at converting it for completing the square.

Step 2: Complete the square

To complete the square, take half of the coefficient of the \( y \) term (which is 6), square it, and add it to both sides. Half of 6 is 3, and \( 3^2 = 9 \). Add 9 to both sides: \( y^2 + 6y + 9 = -8 + 9 \). This transforms the equation into \( (y + 3)^2 = 1 \).

Step 3: Solve the equation

Now solve for \( y \) by taking the square root of both sides of the equation \( (y + 3)^2 = 1 \). This gives \( y + 3 = \pm 1 \). Solve for \( y \) by subtracting 3 from both sides: \( y = -3 \pm 1 \).

Step 4: Simplify the solutions

Calculate the two potential solutions from \( y = -3 + 1 \) and \( y = -3 - 1 \). Thus, you get \( y = -2 \) and \( y = -4 \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra. They are polynomials of degree two, featuring a variable raised to the second power. A typical quadratic equation is written in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. Quadratics can describe various phenomena in mathematics and physics, such as parabolas and projectile motion.
Understanding the standard form of quadratic equations is essential as it helps us identify the properties and behaviors of the equation. The coefficient \( a \) determines the direction and width of the parabola. The term \( b \) affects the placement of the vertex, and \( c \) represents the y-intercept.
In the context of the exercise, the equation \( y^2 + 6y = -8 \) is a specific form of a quadratic equation. It's not initially set to zero, and solving it involves a specific technique known as "completing the square."

Solving Equations

The goal of solving an equation is to find the value(s) of the variable that satisfy it. In the exercise, we're solving the quadratic equation using the "completing the square" method. This method transforms the quadratic equation into a perfect square trinomial, making it easier to solve.
First, make sure all variable terms are on one side of the equation and the constant term on the other side. While this step was erroneously rearranged in the solution, it's crucial in other equations. For the equation at hand, \( y^2 + 6y = -8 \), we didn't need this rearrangement.
Next, to complete the square, add half the coefficient of the linear term squared to both sides. For \( y^2 + 6y \), take half of 6, which is 3, square it to get 9, and add it to both sides, resulting in \( y^2 + 6y + 9 = 1 \). This process enables solving by isolating a binomial square, like \( (y + 3)^2 \).

Algebraic Manipulation

Algebraic manipulation involves rearranging an equation to make it easier to solve. In completing the square, this technique becomes vital. We adjust the initial equation \( y^2 + 6y + 9 = -8 + 9 \) to \( (y + 3)^2 = 1 \), which allows for straightforward solving.
After forming a perfect square, take the square root of both sides to isolate the variable term. From \( (y + 3)^2 = 1 \), find \( y + 3 = \pm 1 \).
This step splits into two separate simple equations: \( y + 3 = 1 \) and \( y + 3 = -1 \). Subtract 3 from both equations to solve for \( y \), resulting in \( y = -2 \) and \( y = -4 \).
Algebraic manipulation is not just about moving numbers around; it's about seeing the equation's structure and knowing how to rearrange it to reveal the solutions.