Completing the square is one of the methods in solving a quadratic equation. A polynomial equation in which one side is a perfect square trinomial is solved by taking square root on each side.

This process of solving the equation is known as Completing the square. We use a binomial to produce a perfect square trinomial.

An equation where one side is perfect square trinomial is solved by taking the square root on each side. If the given equation cannot be factorized then the algebraic expression is written as a square plus another term. The other term is easily found by dividing the coefficient of x by 2 and squaring it.



Consider the Quadratic Equation 

Turn it into this form:

 
 
$ax^{2} + bx + c =0$ 
------->
  $a(x+d)^{2} + e = 0$

The formula for completing the square is given below:
$ax^{2} + bx + c =0$ = $a(x+d)^{2} + e = 0$

where d = $\frac{b}{2a}$ and e = c - $\frac{b^{2}}{4a}$
Given below are the steps to be considered for solving the given expression in completing the square method:
  1. For the highest coefficient in the given term, divide each term by that value in the given expression.
  2. Move the constant to the right hand side.
  3. To find the value needed to make it a perfect square trinomial half the middle term (x), square it and add it to both sides of the equation.
  4. Factor the perfect square trinomial.
  5. Solve by taking square root on each side and make sure to consider both plus and minus results in the solution.

Solved Examples

Question 1: Using Completing the square method solve the equation $3x^{2}+ 2x +9 = 0$
Solution:
 
The coefficient of x2 is 3. Divide each term by 3 on both sides.

$x^{2}$ + $\frac{2}{3}$ x + 3 = 0
        
Move the constant to the right hand side.

$x^{2}$ + $\frac{2}{3}$ x = -3

Half the x- term coefficient, square it and add this value to both sides.

x-term coefficient = $\frac{2}{3}$

Half the x-term coefficient = $\frac{1}{3}$

After Squaring = $\frac{1}{9}$

Add this value to both sides

$x^{2}$ + $\frac{2}{3}$x + $\frac{1}{9}$ = -3 + $\frac{1}{9}$

Simplifying the above

$x^{2}$ + $\frac{2}{3}$x + $\frac{1}{9}$ = -$\frac{26}{9}$

The left side of the expression should be converted to perfect square and by taking square root on both sides

 (x+$\frac{1}{3})^{2}$ = -$\frac{26}{9}$

  x+$\frac{1}{3}$= $\pm\sqrt{\frac{-26}{9}}$

Solve for x:

           x=-$\frac{1}{3}$$\pm\sqrt{\frac{-26}{9}}$

           $x_{1}$ = -$\frac{1}{3}$ + $\frac{1}{3}\sqrt{26i}$

           $x_{2}$ = -$\frac{1}{3}$ - $\frac{1}{3}\sqrt{26i}$
 

Question 2: Using Completing the square method solve the equation $9x^{2}- 7x -9 = 0$
Solution:
 
The coefficient of x2 is 9. Divide each term by 9 on both sides.

$x^{2}$ - $\frac{7}{9}$ x - 1 = 0
        
Move the constant to the right hand side.

$x^{2}$ - $\frac{7}{9}$ x = 1

Half the x- term coefficient, square it and add this value to both sides.

x-term coefficient = -$\frac{7}{9}$

Half the x-term coefficient = -$\frac{7}{18}$

After Squaring = $\frac{49}{324}$

Add this value to both sides

$x^{2}$ - $\frac{7}{9}$x + $\frac{49}{324}$ = 1 + $\frac{49}{324}$

Simplifying the above

$x^{2}$ - $\frac{7}{9}$x + $\frac{49}{324}$ = $\frac{373}{324}$

The left side of the expression should be converted to perfect square and by taking square root on both sides

 (x-$\frac{7}{18}$$)^{2}$ = $\frac{373}{324}$

  x-$\frac{7}{18}$ = $\pm\sqrt{\frac{373}{324}}$

Solve for x:

           x = $\frac{7}{18}$$\pm\sqrt{\frac{373}{324}}$

           $x_{1}$ = $\frac{7}{18}$ + $\frac{1}{18}\sqrt{373}$

           $x_{2}$ = $\frac{7}{18}$ - $\frac{1}{18}\sqrt{373}$