My class is doing a unit on the "completing the square method" which is used to solve quadratic equations. I need help understanding this method. It's not making any sense right now! I have two problems and I need to figure out which number completes the square. Here are the problems:

1: x^2+8x
2: y^2+10y

First one: Take one half of eight, and square it: 16
Second one: take one half of 10, square it.

To understand the completing the square method for solving quadratic equations, let's break it down step by step.

Step 1: Identify the coefficient of the linear term
In your first problem, x^2+8x, the coefficient of the linear term is 8. In the second problem, y^2+10y, the coefficient of the linear term is 10.

Step 2: Take half of the coefficient and square it
In the first problem, take half of 8, which is 4, and square it: 4^2 = 16.
In the second problem, take half of 10, which is 5, and square it: 5^2 = 25.

Step 3: Add the square term inside the parentheses
Now, let's complete the square for each problem:

1: x^2+8x
To complete the square, we need to add the square term inside the parentheses. The square term is 16, which we found in step 2. So we rewrite the equation by adding 16 inside the parentheses:

x^2 + 8x + 16

2: y^2+10y
For this problem, the square term is 25. So we rewrite the equation by adding 25 inside the parentheses:

y^2 + 10y + 25

Step 4: Rewrite the equation
After adding the square term inside the parentheses, rewrite the equation by factoring the square term. This step creates a perfect square trinomial.

1: x^2 + 8x + 16 can be factored as (x + 4)^2
2: y^2 + 10y + 25 can be factored as (y + 5)^2

Step 5: Solve the equation
Now that we have the equations in the form of perfect square trinomials, we can solve them by taking the square root of both sides:

1: (x + 4)^2 = 0 => x + 4 = 0 => x = -4
2: (y + 5)^2 = 0 => y + 5 = 0 => y = -5

So, the solutions to the quadratic equations are x = -4 and y = -5, respectively.

Completing the square method helps you solve quadratic equations by converting them into perfect square trinomials, making them easier to solve.