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# Completing the Square – Explanation & Examples

So far, you’ve learnt how to factorize special cases of quadratic equations using the difference of square and perfect square trinomial method.

These methods are relatively simple and efficient; however, they are not always applicable to all quadratic equations.

In this article, we are going to learn **how to solve all types of quadratic equations** using a simple ** method known as completing the square**. But before that, let’s have an overview of the quadratic equations.

A quadratic equation is a polynomial of second degree usually in the form of f(x) = ax^{2} + bx + c where a, b, c, ∈ R and a ≠ 0. The term ‘a’ is referred to as the leading coefficient, while ‘c’ is referred to as the absolute term of f (x).

Every quadratic equation has two values of the unknown variable usually known as the roots of the equation (α, β). The roots of a quadratic equation can be obtained by factoring the equation.

## What is Completing the Square?

Completing the square is a method of solving quadratic equations that cannot be factorized.

**Completing the square simply means to manipulate the form of the equation so that the left side of the equation is a perfect square trinomial.**

## How to Complete the Square?

To solve a quadratic equation; ax^{2 }+ bx + c = 0 by completing the square.

*The following are the procedures:*

- Manipulate the equation in the form, such that the c is alone on the right side.
- If the leading coefficient a is not equals to 1, then divide each term of the equation by a, such that co-efficient of x
^{2 }is 1. - Add both sides of the equation by the square of half of the co-efficient of term-x

⟹ (b/2a)^{2}.

- Factor the left side of the equation as the square of the binomial.
- Find the square root of both sides of the equation. Apply the rule (x + q)
^{2 }= r, where

x + q= ± √r

- Solve for variable x

### Complete the square formula

In mathematics, completing the square is used to compute quadratic polynomials. Completing the Square Formula is given as: ax^{2 }+ bx + c ⇒ (x + p)^{2 }+ constant.

The quadratic formula is derived using a method of completing the square. Let’s see.

Given a quadratic equation ax^{2 }+ bx + c = 0;

Isolate the term c to right side of the equation

ax^{2 }+ bx = -c

Divide each term by a.

x^{2 }+ bx/a = -c/a

Write as a perfect square

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

(x + b/2a) ^{2}= (-4ac+b^{2})/4a^{2}

(x + b/2a) = ±√ (-4ac+b^{2})/2a

x = – b/2a ±√ (b^{2}– 4ac)/2a

x = [- b ±√ (b^{2}– 4ac)]/2a………. (This is the quadratic formula)

Now let’s solve a couple of quadratic equations using the completing square method.

*Example 1*

Solve the following quadrating equation by completing square method:

x^{2} + 6x – 2 = 0

__Solution__

Transform the equation x^{2} + 6x – 2 = 0 to (x + 3)^{2} – 11 = 0

Since (x + 3)^{2} =11

x + 3 = +√11 or x + 3 = -√11

x = -3+√11

OR

x = -3 -√11

But √11 =3.317

Therefore, x = -3 +3.317 or x = -3 -3.317,

x = 0.317 or x = -6.317

*Example 2*

Solve by completing square x^{2} + 4x – 5 = 0

__Solution__

The standard form of completing square is;

(x + b/2)^{2} = -(c – b^{2}/4)

In this case, b = 4, c = -5. Substitute the values;

So, (x + 4/2)^{2} = -(-5 – 4^{2}/4)

(x + 2)^{2} = 5 + 4

⇒ (x + 2)^{2} = 9

⇒ (x + 2) = ±√9

⇒ (x + 2) = ± 3

⇒ x + 2 = 3, x + 2 = -3

⇒ x = 1, -5

*Example 3*

Solve x^{2} + 10x − 4 = 0

__Solution__

Rewrite the quadratic equation by isolating c on the right side.

x^{2} + 10x = 4

Add both sides of the equation by (10/2)^{2 }= 5^{2 }= 25.

= x^{2 }+ 10x + 25 = 4 + 25

= x^{2 }+ 10x + 25 = 29

Write the left side as a square

(x + 5) ^{2 }= 29

x = -5 ±√29

x = 0.3852, – 10.3852

*Example 4*

Solve 3x^{2} – 5x + 2 = 0

__Solution__

Divide each term of the equation by 3 to make the leading coefficient equals to 1.

x^{2} – 5/3 x + 2/3 = 0

Comparing with the standard form; (x + b/2)^{2} = -(c-b^{2}/4)

b = -5/3; c = 2/3

c – b2/4 = 2/3 – [(5/3)2/4] = 2/3 – 25/36 = -1/36

Therefore,

⇒ (x – 5/6)^{2} = 1/36

⇒ (x – 5/6)= ± √(1/36)

⇒ x – 5/6 = ±1/6

⇒ x = 1, -2/3

*Example 5*

Solve x^{2 }– 6x – 3 = 0

__Solution__

x^{2 }– 6x = 3

x^{2 }– 6x + (-3)^{2 }= 3 + 9

(x – 3)^{2 }= 12

x – 3= ± √12

x = 3 ± 2√3

*Example 6*

Solve: 7x^{2 }− 8x + 3=0

__Solution__

7x^{2 }− 8x = −3

x^{2 }−8x/7 = −3/7

x^{2 }– 8x/7 +(−4/7)^{2 }= −3/7+16/49

(x − 4/7)^{2 }= −5/49

x = 4/7 ± (√7) i/5

(x – 3)^{2 }= 12

x − 3 = ±√12

x = 3 ± 2√3

*Example 7*

Solve 2x^{2} – 5x + 2 = 0

__Solution__

Divide each term by 2

x^{2} – 5x/2 + 1 = 0

⇒ x^{2} – 5x/2= -1

Add (1/2 × −5/2) = 25/16 to both sides of the equation.

= x^{2} – 5x/2 + 25/16 = -1 + 25/16

= (x – 5/4)^{2} = 9/16

= (x – 5/4)^{2} = (3/4)^{2}

⇒ x – 5/4= ± 3/4

⇒ x = 5/4 ± 3/4

x = 1/2, 2

*Example 8*

Solve x^{2}– 10x -11= 0

__Solution__

Write the trinomial as a perfect square

(x^{2} – 10x + 25) – 25 – 11 = 36

⇒ (x – 5)^{2 }– 36 =0

⇒ (x – 5)^{2 }= 36

Find the square roots on both sides of the equation

x – 5 = ± √36

x -5 = ±6

x = −1 or x =11

*Example 9*

Solve the following equation by completing the square

x^{2 }+ 10x – 2 = 0

__Solution__

x^{2 }+ 10x – 2 = 0

⇒ x^{2 }+ 10x = 2

⇒ x^{2 }+ 10x + 25 = 2 + 25

⇒ (x + 5)^{2} = 27

Find the square roots on both sides of the equation

⇒ x + 5 = ± √27

⇒ x + 5 = ± 3√3

x = -5 ± 3√3

*Example 10*

Solve x^{2 }+ 4x + 3 = 0

__Solution__

x^{2 }+ 4x + 3 = 0 ⇒ x^{2 }+ 4x = -3

x^{2} + 4x + 4 = – 3 + 4

Write the trinomial as a perfect square

(x + 2)^{2} = 1

Determine the square roots on both sides.

(x + 2) = ± √1

x= -2+1= -1

OR

x = -2-1= -3

*Example 11*

Solve the equation below using the method of completing the square.

2x^{2} – 5x + 1 = 0

__Solution__

x^{2}−5x/2 + 1/2=0

x^{2 }−5x/2 = −1/2

(1/2) (−5/2) =−5/4

(−5/4)^{2 }= 25/16

x^{2 }− 5x/2 + 25/16 = −1/2 + 25/16

(x – 5/4) ^{2 }= 17/16

Find the square of both sides.

(x – 5/4) = ± √ (17/16)

x = [5 ± √ (17)]/4

*Practice Questions*

Solve the equations below using the method of completing the square.

- 𝑥
^{2}+ 6𝑥 + 5 = 0 - x
^{2}+ 8𝑥 – 9 = 0 - x
^{2}– 6𝑥 + 9 = 0 - 𝑥
^{2}+ 4𝑥 – 7 = 0 - 𝑥
^{2 }– 5𝑥 – 24 = 0 - x
^{2}– 8𝑥 + 15 = 0 - 4x
^{2 }– 4𝑥 + 17 = 0 - 9𝑥
^{2}– 12𝑥 + 13 = 0 - 4𝑥
^{2}– 4𝑥 + 5 = 0 - 4𝑥
^{2}– 8𝑥 + 1 = 0 - x
^{2}+ 4x − 12 = 0 - 10x
^{2}+ 7x − 12 = 0 - 10 + 6x – x
^{2}= 0 - 2x
^{2}+ 8x − 25 = 0 - x
^{2}+ 5x − 6 = 0 - 3x
^{2}− 27x + 9 - 15 − 10x – x
^{2} - 5x
^{2}+ 10x + 15 - 24 + 12x − 2x
^{2} - 5x
^{2}+ 10x + 15

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