Completing the Square method

Example 1:

Solve the x² - 8x + 3 = 0 using completing the square method 


Steps to solve

1. Rearrange the equation by moving +3 to the right-hand side
  x²-8x = -3

2. Take the square of half of the coefficient of x . The coefficient of x is -8 
Half of -8 is - 8/2 = -4
Square of -4 is ( -4 ) ²  = -4x - 4 = + 16

3 .  Add +16 to both sides of the equation
x²- 8x + 16 = -3 + 16
x² - 8x + 16  =  13

Note :  x²- 8x + 16 is now  а  perfect square . 
It can be re - written as x²-8x + 4²
Which is also (x-4 )² = x² -8x + 4²
Check by expanding ( x - 4 ) ( x - 4 ) .

4. Re - write x² - 8x +16 in the perfect square form .
(x-4)²=13


5. Take the square root of both sides .
√(x-4)²= ± √13
Note : Square root will cancel square on the LHS

x - 4  = ±√13

6. Move -4 to the right-hand side 
x = ± √13 + 4 .

7. √13  is an irrational number . Find the  Square  root and solve

x = ±3.606 +4 .

x = + 3.606 + 4  or  -3.606 + 4

=  7.606  or  0.394 .

Answer :  x = 7.606 or 0.394 .

Example 2:

Solve the equation a² + 3a -2 = 0 using
completing the square  method .

Steps to solve
1. Rearrange the equation by moving -2 to the right-hand side ( RHS )
a² + 3a = 2 .

2. Take the square of half of the coefficient of 'a'
The coefficient of 'a' is +3 
Half of +3 is +3/2 
Square of +  is +3/2 is (3/2)² = 9/4

3 .  Add 9/4 to both sides of  the equation
a² + 3a + 9/4  =  2 + 9/4

4  Add  2 + 9/4
(8 + 9)/4 = 17/4

Note that  a² + 3a +9/4  is now  а  perfect square .
It  may  be re - written as  a² + 3a + (3/2)²
a² + 3a + (3/2)² = ( a + 3/2)²

5. Re- write the entire equation in the perfect square form
a² + 3a + 9/4 = 2 + 9/4 becomes;
(a + 3/2)² = 17/4

6  Take the root of both sides or transfer the power of 2 from the left to the right-hand side to become square root 
a + 3/2 = ±√(17/4)
Note the introduction of  ±

7.  Move +3/2 to the right-hand side and solve
 a = ±√((17/4) - 3/2

= ±√4.25  - 1.5

= ± 2.062 - 1.5

= +2.062 -1.5 or -2.062 -1.5

x  =  0.562  or  3.562

Answer :  x = 0.562  or  3.562.0

Example 3:

Solve the equation 3x² - 5x - 1 = 0 using completing the square method .

Steps to solve
1. Divide each term of the equation by the coefficient of x² which is 3.
(3/3)x² -(5/3)x - 7/3 = 0
x² -(5/3) x - 7/3 = 0

Note : The coefficient of x² must be I before other things . 

2. Rearrange the equation by moving (7/3) to the RHS .
x² - (5/3)x = 7/3

3. Take the square of half of the coefficient of x is -3  the coeffici
The coefficient of x is -(5/3)
Half of -(5/3) is -(5/3) multiplied by 1/2 
-(5/3) × ½ = - ⅚
Square of -⅚ = (-⅚)² = 25 /36

4 .  Add 25/26 to both sides of the equation 
x² - (5/3)x + 25/26 = 7/3 + 25/ 36

5. Add 7/3 + 25/36
7/3 + 25/36
84/36 + 25/36 = 109/36

Note : x² + (5/3)x + 25/36 is now a perfect square. It may be re - written as x² - (5/3)x + (5/6)² or (x - 5/6)²

6. Re - write the entire equation in step 4 .
( x - 5 )²  = 109/36

7. Take the root of both sides or transfer the power of 2 from LHS to RHS to become square root .
x - ⅚ = ±√(109/36)

8. Move -⅚ to the RHS . Find the square root of 109/36
x = ±√(109/36) + ⅚
x = ±√3.028 +0.833
= ±1.740 +0.833
= +1.740 + 0.833  or -1.740 + 0.833

Answer : x = 2.573 or -0.907

Example 4:

Solve the equation 3x² - 8x + 2 = 0 using completing the square method.

Steps to solve
1. Divide each term of the equation by the coefficient of x² which is 3
(3/3)x² - (8/3)x + ⅔ = 0
x² - (8/3)x + ⅔ = 0
Note : The coefficient  of x² must be 1 before the other steps. 

2. Rearrange the equation by moving +⅔ to the RHS
x² - (8/3)x =-⅔
Note : The +⅔ becomes -⅔

3. Take the square of half of the coefficient of x. The coefficient of x is  -8/3
Half of -8/3 is -8/3 multiplied by ½
- 8/3 × ½ = -8/6  
Square of -8/6 = ( -8/6 )² = 64/36

4. Add 64/36 to both sides of the equation
x² - (8/3)x + 64/36 = -⅔ + 64/36

5  Add -⅔ and 64/36
-⅔ + 64/36
-24/36 + 64/36 = 40/36
In the simplest form = 10/9
Note: x² - (8/3)x + 64/36 is now a perfect square.
It may be re - written as x² - (8/3)x + ( 8/6 ) ² 
x²- (8/3)x + ( 8/6 )² = ( x - 8/6 ) ²

6. Re - write the entire equation in step 4 it as 
x² - (8/3)x + 64/36 = -⅔ + 64/36
(x - 8/6)² = 10/9

7. Take the root  of  both sides or transfer the power of 2 from the LHS to the RHS to become square root.
x - 8/6 = ±√(10/9)

8  Move -8/6 to the RHS . Find the square root of 10/9 .

x = ±√(10/9) + 8/6
x =  ±√1.111 + 1.333 
x = ± 1.054 + 1.333
x = +1.054 +1.333 
x = 1.054 + 1.333 or -1.054 + 1.333

Answer :  x = 2.387 or 0.279

Example 5:

What must be added to  z² -7z to make the expression a perfect square ?

Steps to solve
1. Take the square of half of the coefficient of z. The coefficient of z is -7
Half of -7 is Square of -7/2
Square of -7/2 is (-7/2)² = 49/4

2. Add 49/4 to the expression . This is what must be added

z² - 7z+ 49/4   - is a  perfect square

Note : The above expression can be re - written as 
z² -7z + ( 7/2)² = ( z - 7/2)²

Answer  = 49/4

Example 6:

Add a term to u² - 1⅗u to make it a  perfect square

Steps to solve

1. Take the square of half of the coefficient of u. The coefficient of u = -1⅗ = -8/5

Half of -8/5  is -8/5 multiplied by 1/2
-8/5 × ½ = -8/10
Square of  -8/10 = (-8/10)² = 64/100

2 .  Add 64/100 to the expression
u² - 1⅗u +64/100
This ( 64/100 ) is what must be added
Note: The above expression  can be re - written as:
u² - 1⅗u + (8/10)² = ( u - 8/10 ) ²

8/10 = ⅘

The expression →(u - 4/5)²

Answer : 64/100 or 16/25 in simplest form

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