Completing the Square Method is a method used in algebra to solve quadratic equations, simplify expressions, and understand the properties of quadratic functions. The method transforms a quadratic equation into a perfect square trinomial, making it easier to solve or analyze.

Algebra is a mathematical science that studies diverse symbols that represent quantities that do not have a set value or amount associated with them but instead vary or change over time in connection to some other feature. In algebra, such symbols are known as variables, and the numbers associated with them are known as coefficients. Here, we will discuss solving quadratic equations using completing the square method with steps and example problems:

What is Quadratic Equation?

A quadratic equation is an algebraic statement of the second degree in x. The quadratic equation is written as ax² + bx + c = 0, with a and b being the coefficients, x being the variable, and c being the constant factor. The coefficient of x² must not be zero (a ≠ 0) for an equation to be classified as a quadratic equation.

The x² term comes first, followed by the x term, and finally the constant term when writing a quadratic equation in standard form. Integral values, rather than fractions or decimals, are typically used to denote the numeric values a, b, and c.

Solving Quadratic Equations using Completing the Square Method

A method or approach for converting a quadratic polynomial or an equation into a perfect square with an additional constant is called completing the Square Method. Using the formula or approach of the complete square, the quadratic equation in the variable x, ax² + bx + c, where a, b and c are the real values ​​except a = 0, can be transformed or converted to a perfect square with an additional constant.

Formula for Completing the Square Method

The completing the square method formula is given by,

ax² + bx + c ⇒ a(x + m)² + n

Where, 

m = b/2a,

n = c – (b²/4a)

Here, m can be any real number and n is a constant.

Completing the Square Method Derivation

Suppose the given quadratic equation is ax² + bx + c. It can be written as,

ax² + bx + c = a (x² + (b/a)x) + c

= a (x² + 2 (bx/2a) + (b/2a)² – (b/2a)²) + c

= a (x² + 2 (bx/2a) + (b/2a)² – b²/4a²) + c

= a (x² + 2 (bx/2a) + (b/2a)²) + (c – b²/4a)

Using the property (a + b)² = a² + b² + 2ab we get,

= a (x + b/2a)² + (c – b²/4a)

Replacing b/2a by a real number m and (c – b²/4a) by a constant value we get,

= a(x + m)² + n

This derives the formula for completing the square method.

Steps for Completing the Square Method in Simple Terms

Here are the steps to solve a quadratic equation using the completing the square method:

Step 1. Start with the quadratic equation:

ax^2 + bx + c = 0

Step 2. Divide by a (if a is not 1) to make the coefficient of x^2 equal to 1.

Step 3. Move the constant term to the other side:

Step 4. Complete the square:

– Take half of the coefficient of x , square it, and add it to both sides:

– Add this value to both sides of the equation:

Step 5. Rewrite the left side as a perfect square:

– The left side of the equation now forms a perfect square trinomial:

Step 6. Solve for x

These steps outline the process for solving quadratic equations by completing the square.

Sample problems on Solving Quadratic Equations using Completing the Square Method

Question 1: Use completing the square method to solve: x² + 4x – 21 = 0.

Solution:

We have, a = 1, b = 4 and c = –21.

Find the value of m and n.

m = 4/2 = 2

n = –21 – (16/4) = –21 – 4 = –25

So, the equation is solved as,

(x + 2)² – 25 = 0

x + 2 = ±5

x = 3, –7

Question 2: Use completing the square method to solve: x² + 10x + 21 = 0.

Solution:

We have, a = 1, b = 10 and c = 21.

Find the value of m and n.

m = 10/2 = 5

n = 21 – (100/4) = 21 – 25 = –4

So, the equation is solved as,

(x + 5)² – 4 = 0

x + 5 = ±2

x = –3, –7

Question 3: Use completing the square method to solve: x² + 6x – 27 = 0.

Solution:

We have, a = 1, b = 6 and c = –27.

Find the value of m and n.

m = 6/2 = 3

n = –27 – (36/4) = –27 – 9 = –36

So, the equation is solved as,

(x + 3)² – 36 = 0

x + 3 = ±6

x = 3, –9

Question 4: Use completing the square method to solve: x² + 12x – 13 = 0.

Solution:

We have, a = 1, b = 12 and c = –13.

Find the value of m and n.

m = 12/2 = 6

n = –13 – (144/4) = –13 – 36 = –49

So, the equation is solved as,

(x + 6)² – 49 = 0

x + 6 = ±7

x = 1, –13

Question 5: Use completing the square method to solve: x² + 20x + 19 = 0.

Solution:

We have, a = 1, b = 20 and c = 19.

Find the value of m and n.

m = 20/2 = 10

n = 19 – (400/4) = 19 – 100 = –81

So, the equation is solved as,

(x + 10)² – 81 = 0

x + 10 = ±9

x = –1, –19

Question 6: Use completing the square method to solve: x² + 6x – 16 = 0.

Solution:

We have, a = 1, b = 6 and c = –16.

Find the value of m and n.

m = 6/2 = 3

n = –16 – (36/4) = –16 – 9 = –25

So, the equation is solved as,

(x + 3)² – 25 = 0

x + 3 = ±5

x = 2, –8

Question 7: Use completing the square method to solve: x² – 4x – 12 = 0.

We have, a = 1, b = –4 and c = –12.

Find the value of m and n.

m = –4/2 = –2

n = –12 – (16/4) = –12 – 4 = –16

So, the equation is solved as, 

(x – 2)² – 16 = 0

x – 2 = ±4

x = 6, –2