**Mathematics** helps us model physical phenomena. When we throw a ball horizontally after following parabolic paths, it falls. This **parabolic path** can be modelled using a quadratic function, as a graph of a quadratic equation gives **parabolic paths.**

** **

Similarly, **golf balls** follow a parabolic path, so for accuracy, we need to understand quadratic functions.

For a better understanding of a quadratic equation, we need to solve it.

There are three methods that are commonly used to find the solution to a quadratic equation.

- Factorization method

- Completing the square method

- Quadratic Formula.

In this topic, we will discuss the completing square method. This method is not only used to find the solution of a quadratic equation but also to find the turning point of a parabolic function. This method is also used to solve the integration method in a few cases.

This method is very important for various mathematics exams, e.g., math GCSE 2022 edexcel, Edexcel Pearson, math p1, and O-level and A-level. You can check this topic in Maths p1 past papers and Edexel papers p1.

y = (x+2)^{2}+4 has a turning point of (-2,4). The turning point is the maximum or minimum point of a parabolic graph.

Now we take various questions to understand the completing square method.

**Example 1**

**Solve 2x ^{2}+8x+4=**

**0 by using completing square method.**

solution

In the given equation, the coefficient of x^{2} is 2.

** Dividing both sides by 2, as for the completing square method, the coefficient of x ^{2} must be 1.**

** we get**

** x ^{2}+4x+2=**

**0**

** Subtracting 2 from both sides, we get. **

** ** x^{2}+4x+2-2=0-2

x^{2}+4x=-2 ** ——————— (1)**

** **Now focus on the coefficient of x, which is 4. To completing square method, we have to multiply 4 by 1/2; we get 4(1/2) = 2. Taking the square of 2. Adding 2^{2} in both sides of (1), we get

x

^{2}+4x+2^{2}=-2 +2^{2}

(x+2)

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

^{2}=a^{2}+2ab+b2, Remember this is the completing the square formula(x+2)

^{2}=2Taking square root of both sides

x+2=±√2

Subtracting 2 from both sides.

x=-2 ±√2

Example 2

Use Completing square method to find the turning point of parabola function y = x

^{2}+4x+2

Solution

y = x

^{2}+4x+2 (2)Multiplying the coefficient of 4 by 1/2, we get 2. Adding and subtracting 2

^{2}on the right sides of (2)we get

y = x

^{2}+4x+2^{2}-2^{2}+2y=(x+2)

^{2}–4 +2y=(x+2)

^{2}-2 completing square formHence, turning point of parabola (2) is (-2, -2).

We use https://www.desmos.com/calculator for graphing of equation (2).

You can see that turning point on graph is (-2,-2).

Example 3

Find the Turning point of y=-3x^{2}+6x+1

solutiony=-3x

^{2}+6x+1 (3)Taking common from the first two terms, as for completing the square method, it is essential that the coefficient of x

^{2}be 1.y=-3(x

^{2}-2x)+1Coefficient of x is 2

and half of 2 is 1, so adding and subtracting 1

^{2}, we gety=-3(x

^{2}-2x+1^{2}-1^{2})+1Separating of –1

^{2}from the bracket gives +3, as -3 outside the bracket will be multiplied by –1^{2}we get

y=-3(x

^{2}-2x+1)+3+1y=-3(x-1)

^{2}+4 as a^{2}-2ab+b^{2}=(a-b)^{2}So, turning point of equation (3) will be (1,4).

Graphing of equation (3), clearly shows the proof of validity of completing square method.

Now we discuss the more completing square method examples.

Example 4

Solution:

(Subtracting constant value ‘6’ on both sides)

half of 5 is 5/2, and square of (5/2) is (5/2)

^{2}(Writing in the form of a

^{2}+2ab+b^{2}by adding value of ‘b^{2}’ on both sides)(Writing in form of a

^{2}– b^{2})or (Taking each factor equal to zero)

Solution set: {2,3}

Example 5

Solution:

(Adding constant value ‘1’ on both sides)

Coefficient of x is 1, and half of 1 is 1/2. Square of 1/2 gives (1/2)

^{2}=1/4. Adding 1/4 on both sides of equation(Writing in constant value in square form)

(Writing in form of ‘a

^{2}– b^{2}‘)(Writing the factors (a+b) (a-b) )

or (Taking each factor equal to zero)

Example 6

Solution:

Example 7

Solution:

We discuss more completing the square questions and answers.

Example 8

Solution:

Example 9

Solution:

Our TARGET IS Solving quadratic equations by completing the square

We discuss more completing the square practice questions.

Example 10

Solution:

Our target is the solving quadratic equation by completing the square

Example 11

Solution:

Our target is the solving quadratics by completing the square

Now we shall try to solve the integration problems, which can be solved using completing square method.

Now remember the formula for integration

∫ (1/(x

^{2}+a^{2}) dx = 1/a tan^{ –1}(x/a) +c (A)Example 12

∫ (1/(x

^{2}+4x +1) dx ——-(B )completing the square method gives x

^{2}+4x+5= (x+2)^{2}+1Put in ( B) and using (A ) gives

∫ (1/[(x+2)

^{2}+1^{2}]dx =tan^{ –1}(x+2)+CSo, you can see that various integration problems can be solved using completing square method.