Address by completing the square may take a little longer to do than problems by factoring. However, the steps are simple.

Prior to showing examples, you need to understand what is a perfect square trinôme

Same binomial × binomial = perfect square trinôme

(x + 4) × (x + 4) = x 2 + 4 x 4 x + 16 = x 2 + 8 x + 16

(x + a) × (x + a) = x 2, a2 = x 2 + ax + ax + have + a2

Important observation:

What is the relationship between the coefficient of the second term and the last term?

X 2 + x 8 + 16, coefficient of the second term is 8 and the last term is 16

(8/2) 2 = 42 = 16

For, 2 x + have + a2, the coefficient of the second term is 2 and the last term is a2

(2 a, 2) 2 = a2

Then, what is the relationship?

The last term is obtained by dividing the coefficient of the second term by 2 and the result of the quadrature

Now, suppose that you have x 2 + 20 x and you want to find the last term to make the thing a trinôme square perfect

Simply do so (2-20) 2 = 102. Square perfect trinôme is x 2 + 20 x + 102 = (x + 10) × (x + 10)

Example # 1:

Address by completing the square 2 x + 6 x 8 = 0

x 2 + x 6 + 8 = 0.

Subtract 8 from both sides of the equation.

2 x 6 x + 8-8 = 0-8

x 2 + 6 x = – 8

To complement the place, always do the following 24 hours a day, 365 days a year. It will never change when you solve by filling the place!

You are basically looking for a term added to 2 x + 6 x that will make a perfect square trinôme.

To this end, obtain the coefficient of the second term, divide by 2 and raise to the second power.

The second term is 6 x and the coefficient is 6.

6/2 = 3 and after 3 squaring, we get 32

x 2 + 6 x = – 8

Add 32 to both sides of the equation above

x 2 + 6 x + 32 =-8 + 32

x 2 + 6 x + 32 =-8 + 32

(x + 3) 2-8 + 9 =

(x + 3) 2 = 1.

Take the square root of both sides

v ((x + 3) 2) = v(1)

x + 3 = ±1.

When x + 3 = 1, x = – 2

When x + 3 = – 1, x = – 4

Example # 2:

Address by completing the square 2 x + 8 = 0 instead of 2 x + x 6 + x 6 + 8 = 0

The second term, this time is x-6 and the coefficient is – 6.

-6/2 = – 3 and after squaring-3, we get (-3) 2 = 9.

x 2 + 6 x = – 8

Add 2 (-3) on both sides of the equation above

x-2 + 6 x + 2 (-3) =-8 + 2-(3)

-(x + 3) 2-8 + 9 =

-(x + 3) 2 = 1.

Take the square root of both sides

v ((x + -3) 2) = v(1)

-x + 3 = ±1.

When-x + 3 = 1, x = 4

Where x +-3 = – 1, x 2

Example # 3:

Address by completing the square of 3 x 2 + 8 x +-3 = 0.

3 x 2 + 8 x +-3 = 0.

Everything is divided by 3. Always do so when the coefficient of the first term is not 1

(3/3) x 2 + x (3/8) + – 3/3 = 0 / 3

x 2 + x (3/8) + 1 = 0.

Add 1 on both sides of the equation.

x 2 + x (3/8) + 1 + 1 = 0 + 1

x 2 + x (3/8) = 1

The second term is x (3/8) and the coefficient is 8/3.

8/3 ÷ 2 = 8 / 3 × 1/2 = 8 / 6 and after 8/6 square, we obtain 2 (8/6)

x 2 + x (3/8) = 1

Add 2 (8/6) on both sides of the equation above

x 2 + x (8/3) + (8/6) 2 = 1 + 2 (8/6)

(x + 8/6) 2 = 1 + 64/36

(x + 8/6) 2 = 36/36 + 64/36 = (36 + 64) / 36 = 100/36

Take the square root of both sides

v ((x_+_8/6) 2) = v(100/36)

x + 8/6 = ±10/6.

x + 8/6 = 10/6

x = 10 / 6-8/6 = 2 / 6 = 1/3.

x + 8/6 = – 10/6.

x = – 10/6-8/6 = – 18/6 = – 3

To resolve by completing the square may become quickly hard as shwon in example # 3