# Completing the Square: Definition, Formula / Steps, Examples

## What is Completing the Square

Completing the square is one of method to solve quadratic form. Base concept that used in this method are:

(a+b)2 = a2 + 2ab +b2

(a-b)2 = a2 – 2ab +b2

Sometimes, if quadratic form cannot be solved (can be solve hardly) by factorization method, it can use completing the square method.

It is used to convert quadratic form (ax2+bx+c=0) into another form a(x-h)2+k where h and k is any values.

ax2+bx+c to be a(x-h)2+k

There are some cases that use completing the square, they are solving, deriving, graphing quadratic function (formula), finding Laplace transform, and evaluating integral.

## Completing the Square Formula / Steps

Generally, converting quadratic form to be a(x-h)2+k using completing the square is not easy if it is done manually. But there is formula to make it easier.

ax2+bx+c = a(x-h)2+k

where

• h = -b/2a
• k = c – (a.h2) = c – (b2/4a)

there is special case if a = 1

x2+bx+c = (x-h)2+k

where

• h = -b/2
• k = c – (h2) = c – (b2/4)

In another words, there is steps to describe how to use completing the square method:

1. Making coefficient of ax2 is 1 (divide ax2 by a)
2. Moving constant to the right side
3. Completing the square on the left side (it is not balance yet between left and right sides) and balancing the left and right side by adding same value
4. Taking square root on the both sides
5. Subtracting the number on the right sides so the left side become x.

What is different between steps and formula?

The both methods are used to completing the square. It can be used depend on each person which one is easier to solve the problem.

## Completing the Square Examples

1. Solve x2 + 4x + 1 = 0

For the first, use the steps to understand easily.

Step 1: x2 + 4x + 1 = 0 (a =1)

Step 2: x2 + 4x = -1

Step 3: x2 + 4x + 4 = -1 +4 then (x+2)2 = 3

Step 4: √(x+2)2 = ±√3 → x+2 = √3

Step 5: x = ±√3 -2

So, the solution of x2 + 4x + 1 = 0 is x = ±√3 -2.

2. Solve x2 – 4x + 3 = 0

Using formula

x2+bx+c = (x-h)2+k

then

x2 – 4x + 3 = 0

h = (-b/2a) = (-(-4)/2.1) = 2

k = c – h2 = 3 – 22 = 3 – 4 = -1

x2 – 4x + 3 = 0

→ (x – 2)2 – 1 = 0

→      (x – 2)2  = 1

→          x – 2 = ± √1

→               x  = ± √1 + 2

→               x  = ± 1 + 2

So the solutions are

x = 1 + 2 = 3

or

x = -1 + 2 = 1

3. Solve 2x2 – 5x + 3 = 0

2x2 – 5x + 3 = 0 (divide 2)

→    x2 – (5/2) x + 3/2 = 0

→               x2 – (5/2)x = -3/2

→x2 – (5/2)x + 25/16 = -3/2 + 25/16

→               (x – (5/4))2 = 1/16

→                  x – (5/4) = ± √(1/16)

→                              x = ±( ¼) +(5/4)

So the solutions are

x = (¼)+ (5/4) = 6/4 = 3/2

or

x = – (¼) +(5/4) = 4/4 = 1