# Completing the square – getting rid of the square (difficult)

When it isn’t obvious how to get rid of the square

Complete the following square:

Remember the above picture

**NOTE:**

- The coefficient (or number) in front of the `x` must be a 1
- The next square must be ½ of the next term
- Multiply out into all of the boxes

This method uses a visual model to solve a quadratic equation

**For example:**

`x^2+2x-8=0`

Visually `x^2` can look like an area of a table:

The `2x` can be represented by:

And this will therefore complete the last square as:

**NOTE:**

This is the same as `(x+1)^2`

At the moment if you add each area together you get:

`x^2+1x+1x+1times1`

`=x^2+2x+1`

Originally we had `x^2+2x-8=0`

We now have `x^2+2x+1=0`

Always plot this on a number line

The number line will help you remember

Original number `-` New number

`-8-1=-9` (we need to `-9` )

So `x^2+2x-8=0`

Is the same as `(x+1)^2-9=0`

Which can now be solved

`(x+1)^2-9=0`

`(x+1)^2=9`

`x+1=+-sqrt9`

(Don't forget the root of anything can be `+` or `-`)

`x+1=+-3`

`x=-1+-3`

`x=-1+3\ \ \ or\ \ \ x=-1-3`

`x=2\ \ \ or\ \ \ x=-4`

## Now check

`x^2+2x-8=0`

If `x=2` `2^2+2times2-8=0` Which is correct

If `x=-4` `(-4)^2+2times(-4)-8=0`

`16-8-8=0` Which is correct

(If they don't add up to zero you can be assured that it is wrong)

**Answer:**

The roots of `x^2+2x-8=0` are `x=2` and `x=-4`

# Completing the square – getting rid of the square (difficult)

When it isn’t obvious how to get rid of the square

Complete the following square:

Remember the above picture

**NOTE:**

- The coefficient (or number) in front of the `x` must be a 1
- The next square must be ½ of the next term
- Multiply out into all of the boxes

This method uses a visual model to solve a quadratic equation

**For example:**

`x^2+2x-8=0`

Visually `x^2` can look like an area of a table:

The `2x` can be represented by:

And this will therefore complete the last square as:

**NOTE:**

This is the same as `(x+1)^2`

At the moment if you add each area together you get:

`x^2+1x+1x+1times1`

`=x^2+2x+1`

Originally we had `x^2+2x-8=0`

We now have `x^2+2x+1=0`

Always plot this on a number line

The number line will help you remember

Original number `-` New number

`-8-1=-9` (we need to `-9` )

So `x^2+2x-8=0`

Is the same as `(x+1)^2-9=0`

Which can now be solved

`(x+1)^2-9=0`

`(x+1)^2=9`

`x+1=+-sqrt9`

(Don't forget the root of anything can be `+` or `-`)

`x+1=+-3`

`x=-1+-3`

`x=-1+3\ \ \ or\ \ \ x=-1-3`

`x=2\ \ \ or\ \ \ x=-4`

## Now check

`x^2+2x-8=0`

If `x=2` `2^2+2times2-8=0` Which is correct

If `x=-4` `(-4)^2+2times(-4)-8=0`

`16-8-8=0` Which is correct

(If they don't add up to zero you can be assured that it is wrong)

**Answer:**

The roots of `x^2+2x-8=0` are `x=2` and `x=-4`