 # Completing the Square Calculator

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## Completing Square Calculator

Completing Square Calculator is used to solve the quadratic equation by the completing square technique and also make the standard form of the completing square of the algebraic expression.

## What is Completing Square and how it works?

Completing the square is a mathematical method that makes them easy to solve those quadratic equations which are not factorized easily. It converts the equation into a perfect square with the help of this we can easily solve any quadratic equation.

Step to generate the completing square if the quadratic equation in the form of ax2 + b x +c =0.

• Firstly, transform the constant term “c” other side of the equation by replacing its sign as ax2 + b x = −c.
• If the coefficient of “x2” is exit in the equation then divide the whole equation to make the coefficient free.
• Multiply the coefficient of “x” with the “1/2”.
• Added the square of the above term getting after the multiplication of “1/2” on both sides.
• The left side can compress into the squares of the terms and the right side simplify by any algebraic expression.
• Finally, take the square root on both sides and simplify by using any algebraic technique.

## How to solve the problems of completing squares?

Here we solved the different examples to understand the idea of completing a square.

Example 1:

Evaluate the equation using the completing square technique.

2 x2 − 8 x - 10 = 0

Solution:

Step 1:

First, divide with the “2” to make the “x2” coefficient free.

(2/2) x2 – (8/2) x – (10/2) = 0

x2 − 4 x - 5 = 0

Step 2:

Identify the constant term and take it to the other side.

x2 − 4 x - 5 = 0

x2 − 4 x = 5

Step 3:

Take the coefficients of “x” without the sign and multiply with the “1/2” and simplify.

4 × (1/2) = 4 × ½ = 2

Step 4:

Now adding the square of the above finding number on both sides.

x2 − 4x + (2)2 = (2)2 + 5

(x)2 − 4x + (2)2 = 4 + 5

(x)2 − 4x + (2)= 9

(x−2)2 = 9

Step 5:

Now take the square root on both sides and simplify.

(x−2) = ± 3

x = 2 ± 3

• x = 2 + 3

x = 5

• x = 2 - 3

x = -1

Now, the solution of “x”.

x = {− 1, 5}.

Example 2:

Evaluate the equation using the completing square technique.

x2 − 2 x + 5 = 0

Solution:

If the term “x2” in the equation is already free with any number then proceed by the second step.

Step 1:

First, identify the constant term and take it to the other side.

x2 − 2 x + 5 = 0

x2 − 2 x = -5

Step 2:

Take the coefficients of “x” without the sign and multiply with the “1/2” and simplify.

2 × (1/2) = 2 × ½ = 1

Step 3:

Now adding the square of the above finding number on both sides.

x2 − 2x + (1)2 = (1)2 - 5

(x)2 − 2x + (1)2 = 1 - 5

(x)2 − 2x + (1)2 = - 4

(x−1)2 = -4

Step 4:

Now take the square root on both sides and simplify.

(x−1) = ±2i

x = 1 ± 2i

Now, the solution of “x”.

x = {1 − 2i, 1 + 2i}.

Example 3:

Evaluate the equation using the completing square technique.

x2 + 4 x - 12 = 0

Solution:

If the term “x2” in the equation is already free with any number then proceed by the second step.

Step 1:

First, identify the constant term and take it to the other side.

x2 + 4 x - 12 = 0

x2 + 4 x = 12

Step 2:

Take the coefficients of “x” without the sign and multiply with the “1/2” and simplify.

4 × (1/2) = 4 × ½ = 2

Step 3:

Now adding the square of the above finding number on both sides.

x2 + 4x + (2)2 = (2)2 + 12

(x)2 + 4x + (2)2 = 4 + 12

(x)2 + 4x + (2)2  = 16

(x+2)2 = 16

Step 4:

Now take the square root on both sides and simplify.

(x+2) = ± 4

x = -2 ± 4

• x = -2 + 4

x = 2

• x = -2 - 4

x = -6

Now, the solution of “x”.

x = {− 6, 2}.

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