In this method,convert any quadratic equation to the form ${(x + a)^2 – b^2 = 0}$ or ${x^2 – b^2 = 0}$

If the coefficient of ${x^2}$ is a perfect square
How to convert ${x^2 + 4x – 5 = 0}$  to the form  ${(x + a)^2  – b^2  = 0}$.

 


Consider a square with side $x$ and a rectangle with sides $4$ and $x$
So, the sum of area of these two objects will be:

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(-)


${x^2  + 4x =  (x^2  +(4x)/2 ) + (4x)/2}$

=  ${x^2  + 2x + 2x}$
=  ${(x + 2) x + 2 × x}$
=  ${(x + 2) x + 2 × x + 2 × 2 – 2 × 2}$
=  ${(x + 2) x + (x + 2) × 2 – 2 × 2}$
=  ${(x + 2) (x + 2) – 2^2}$
=  ${(x + 2)^2  – 2^2}$

So,    ${x^2  + 4x – 5 =  (x + 2)^2  – 4 – 5 = (x + 2)^2 – 3^2}$

So, ${x^2 + 4x – 5 = 0}$ can be written as ${(x + 2)^2 – 9 = 0}$ by this process of completing the square. 
This is known as the method of completing the square.

If the coefficient of ${x^2}$  is not a perfect square.
Method 1: 

 

Consider the equation ${3x^2 – 5x + 2 = 0}$. 
Since the coefficient of ${x^2}$ is not a perfect square,multiply the equation throughout by 3 to get a perfect square of 3;
${9x^2  – 15x + 6 =  0}$
${9x^2  – 15x }$ = ${(3x)^2 - (5/2)3x}$ ${-(5/2)3x}$
				=${3x}$${(3x-5/2)}$${-(5/2)3x}$
Add and Subract ${(5/2)^2}$ from the above equation.
    =${3x}$${(3x-5/2)}$${-(5/2)3x}$  +${(5/2)^2}$ - ${(5/2)^2}$ 
	=${3x}$${(3x-5/2)}$-${5/2}$${(3x-5/2)}$ -${(5/2)^2}$
	=${(3x-5/2)}$${(3x-5/2)}$-${(5/2)^2}$
	=${(3x-5/2)^2}$-${(5/2)^2}$
	
So, ${9x^2  – 15x + 6 }$ = ${(3x-5/2)^2}$-${25/4+6}$
			=${(3x-5/2)^2}$-${1/4}$ =0
			
i.e ${(3x-5/2)}$-${1/2 =0}$ or  ${(3x-5/2)}$+${1/2 =0}$

i.e ${x=1}$ or ${x=2/3}$


Method 2
Consider the equation ${3x^2–5x+2=0}$. 
Divide the equation by 3 to make it a perfect square.
Therefore ${3x^2–5x+2=0}$= ${x^2–5/3x+2/3=0}$
${x^2–5/3x}$= ${x^2–1/2(5/3x)-1/2(5/3x)}$
			=${x(x-5/6)}$-${5/6x}$
Add and subract ${(5/6)^2}$
Therefore the equation becomes
			=${x(x-5/6)}$-${5/6x}$+${(5/6)^2}$-${(5/6)^2}$
			=${x(x-5/6)}$ -${5/6}$${(x-5/6)}$-${(5/6)^2}$
			=${(x-5/6)}$${(x-5/6)}$ -${25/36}$
			=${(x-5/6)^2}$-${25/36}$
			
${x^2–5/3x+2/3}$=${(x-5/6)^2}$-${25/36+2/3}$
                = ${(x-5/6)^2}$-${(25-24)/36}$
				=${(x-5/6)^2}$-${(1/6)^2}$

Therefore the equation ${3x^2–5x+2=0}$ is same as ${(x-5/6)^2}$-${(1/6)^2=0}$
			Since	${a^2-b^2}$=${(a-b)(a+b)}$
Therefore	${(x-5/6)}$-${(1/6)=0}$ or ${(x-5/6)}$+${(1/6)=0}$
			${x=1}$ or ${x=2/3}$
			
Example:Solve the equation given ${2x^2  – 5x + 3 = 0}$ by the method of completing the square.
Solution  
     Divide the equation by coefficient of ${x^2}$ to make it complete square
	${2x^2  – 5x + 3}$=${x^2  – 5/2x + 3/2}$
	
	${x^2  – 5/2x + 3/2}$=${(x-5/4)^2 -(5/4)^2+3/2}$=${(x-5/4)^2 -1/16}$
(Hint: ${a^2 -2ab+b^2=(a-b)^2}$ , therefore, ${a^2-2ab=(a-b)^2-b^2}$)

Therefore ${2x^2  – 5x + 3 = 0}$ can be written as= ${(x-5/4)^2 -1/16=0}$
${(x-5/4)^2 -(1/4)^2=0}$
${(x-5/4)+(1/4)=0}$ or ${(x-5/4) -(1/4)=0}$
${x-1=0}$ or ${x-6/4=0}$
${x=1}$ or ${x=3/2}$


HomeWork: Solve above equation i.e ${3x^2–5x+2=0}$ by method 1 i.e making the coefficient of 
${x^2}$ a perfect square by multiplying it by coefficient of ${x^2}$

Consider the following situation:
The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?

Solution:let her present age (in years) be ${x}$.
Then the product of her ages two years ago and four years from now is ${(x – 2)(x + 4)}$.
Therefore,${(x – 2)(x + 4) = 2x + 1}$
i.e., ${x^2 + 2x – 8 = 2x + 1}$
i.e.,${x^2 – 9 = 0}$
So, Sunita’s present age satisfies the quadratic equation ${x^2 – 9 = 0}$.
We can write this as ${x^2 = 9}$. Taking square roots, we get ${x = 3}$ or ${x = – 3}$.
Since the age is a positive number, ${x = 3}$.
So, Sunita’s present age is 3 years.

Now consider the quadratic equation ${(x + 2)^2 – 9 = 0}$. To solve it, we can write it as ${(x + 2)^2 = 9}$. Taking square roots, we get ${x + 2 = 3}$ or ${x + 2 = – 3}$.
Therefore, ${x = 1}$ or ${x = –5}$
So, the roots of the equation ${(x + 2)^2 – 9 = 0}$ are 1 and – 5.