Question #8950d

`y=x^2-5x+101`

To complete the square, your goal is to make a perfect trinomial. For example:

`x^2+2x+1=(x+1)^2`

To make things easier, let's add some parentheses to the equation.

`y=(x^2-5x)+101`

You need to add a constant into the parentheses to create a perfect trinomial. To find this constant, use this formula:

`(b/2)^2`

The `b` comes from the standard form of a quadratic equation `y=ax^2+bx+c`. In this case, that means your `b` is `-5`.

`(-5/2)^2`

`(-2.5)^2`

`6.25`

This is where things can get a little confusing. Because you are adding a number on one side of the equation, and not the other side, you need to balance it out on the same side.

`y=(x^2-5x+6.25)+101-6.25`

As you can see, you've created your perfect trinomial while also not unbalancing your equation. Now you can factor and simplify!

`y=(x-2.5)^2+101-6.25`

`y=(x-2.5)^2+94.75`

`x^2-5x+101`

To complete the square, we need to find a value that makes `x^2-5x` a perfect square. To make it easier to visualize, I like to move the other component (`101`) to the other side of the equation.

`-101=x^2-5x`

To solve for the missing component, we need to follow these steps:

1) take the middle term, `-5` and divide by `2`

`(-5)/2=-5/2`

2) square this solution

`(-5/2)^2 = 25/4`

Now, let's add this to the equation. REMEMBER in an equation, we can add whatever we want, but we must also add it to the other side:

`-101 + 25/4 = x^2-5x+25/4`

`-379/4 = (x-5/2)^2`

`(x-5/2)^2+379/4`

Now we have our solution!

`y = (x-5/2)^2+379/4`