"**Completing the Square**" is where we ...

... take a Quadratic Equation like this: | and turn it into this: | |

ax^{2} + bx + c = 0 | a(x+d)^{2} + e = 0 |

For those of you in a hurry, I can tell you that: d = *b***2a**

and:e = c − *b ^{2}*

**4a**

But if you have time, let me show you how to "**Complete the Square**" yourself.

## Completing the Square

Say we have a simple expression like x^{2} + bx. Having x twice in the same expression can make life hard. What can we do?

Well, with a little inspiration from Geometry we can convert it, like this:

As you can see x^{2} + bx can be rearranged * nearly* into a square ...

... and we can **complete the square** with (b/2)^{2}

In Algebra it looks like this:

x^{2} + bx | + (b/2)^{2} | = | (x+b/2)^{2} |

"Complete the Square" |

So, by adding (b/2)^{2} we can complete the square.

The result of (x+b/2)^{2} has x only **once**, which is easier to use.

## Keeping the Balance

Now ... we can't just * add* (b/2)

^{2}without also

*it too! Otherwise the whole value changes.*

**subtracting**So let's see how to do it properly with an example:

Start with: | |

("b" is 6 in this case) | |

Complete the Square: | |

( | |

Simplify it and we are done. | |

The result:

x^{2} + 6x + 7 = (x+3)^{2} − 2

And now x only appears once, and our job is done!

## A Shortcut Approach

Here is a method you may like, it isquick when you get used to it.

First think about the result we want: (x+d)^{2} + e

After expanding (x+d)^{2} we get: x^{2} + 2dx + d^{2} + e

Now see if we can turn our example into that form to discover d and e.

### Example: try to fit x^{2} + 6x + 7 into x^{2} + 2dx + d^{2} + e

Now we can "force" an answer:

- We know that 6x must end up as 2dx, so
**d****must be 3** - Next we see that 7 must become d
^{2}+ e = 9 + e, so**e****must be −2**

And we get the same result (x+3)^{2} − 2 as above!

Now, let us look at a useful application: solving Quadratic Equations ...

## Solving General Quadratic Equations by Completing the Square

We can complete the square to **solve** a Quadratic Equation (find where it is equal to zero).

But a general Quadratic Equation may have a coefficient of a in front of x^{2}:

ax^{2} + bx + c = 0

To deal with that we divide the whole equation by "a" first, then carry on:

x^{2} + (b/a)x + c/a = 0

## Steps

Now we can **solve** a Quadratic Equation in 5 steps:

**Step 1**Divide all terms by**a**(the coefficient of**x**).^{2}**Step 2**Move the number term (**c/a**) to the right side of the equation.**Step 3**Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.

We now have something that looks like (x + p)^{2} = q, which can be solved this way:

**Step 4**Take the square root on both sides of the equation.

**Step 5**Subtract the number that remains on the left side of the equation to find**x**.

## Examples

OK, some examples will help!

### Example 1: Solve x^{2} + 4x + 1 = 0

**Step 1** can be skipped in this example since the coefficient of x^{2} is 1

**Step 2** Move the number term to the right side of the equation:

x^{2} + 4x = -1

**Step 3** Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation.

(b/2)^{2} = (4/2)^{2} = 2^{2} = 4

x^{2} + 4x + 4 = -1 + 4

(x + 2)^{2} = 3

**Step 4** Take the square root on both sides of the equation:

x + 2 = ±√3 = ±1.73 (to 2 decimals)

**Step 5** Subtract 2 from both sides:

x = ±1.73 – 2 = -3.73 or -0.27

And here is an interesting and useful thing. At the end of step 3 we had the equation: (x + 2) It gives us the |

### Example 2: Solve 5x^{2} – 4x – 2 = 0

**Step 1** Divide all terms by 5

x^{2} – 0.8x – 0.4 = 0

**Step 2** Move the number term to the right side of the equation:

x^{2} – 0.8x = 0.4

**Step 3** Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2)^{2} = (0.8/2)^{2} = 0.4^{2} = 0.16

x^{2} – 0.8x + 0.16 = 0.4 + 0.16

(x – 0.4)^{2} = 0.56

**Step 4** Take the square root on both sides of the equation:

x – 0.4 = ±√0.56 = ±0.748 (to 3 decimals)

**Step 5** Subtract (-0.4) from both sides (in other words, add 0.4):

x = ±0.748 + 0.4 = -0.348 or 1.148

## Why "Complete the Square"?

Why complete the square when we can just use the Quadratic Formula to solve a Quadratic Equation?

Well, one reason is given above, where the new form not only shows us the vertex, but makes it easier to solve.

There are also times when the form **ax ^{2} + bx + c** may be part of a

**larger**question and rearranging it as

**a(x+**makes the solution easier, because

*d*)^{2}+*e***x**only appears once.

For example "x" may itself be a function (like *cos(z)*) and rearranging it may open up a path to a better solution.

Also Completing the Square is the first step in the Derivation of the Quadratic Formula

Just think of it as another tool in your mathematics toolbox.

364, 1205, 365, 2331, 2332, 3213, 3896, 3211, 3212, 1206

### Footnote: Values of "d" and "e"

How did I get the values of **d** and **e** from the top of the page?

Start with | |

Divide the equation by a | |

Put c/a on other side | |

Add (b/2a) to both sides^{2} | ^{} |

"Complete the Square" | ^{} |

Now bring everything back... | |

... to the left side | |

... to the original multiple a of x^{2} |

And you will notice that we have:a(x+d)^{2} + e = 0

Where:d = *b***2a**

and:e = c − *b ^{2}*

**4a**

Just like at the top of the page!

Quadratic Equations Factoring Quadratics Graphing Quadratic Equations Real World Examples of Quadratic Equations Derivation of Quadratic Equation Quadratic Equation Solver Algebra Index