Completing the Square and The Square Root Method

 

The Square Root Property


If we have

        x2  =  k 
then we can take the square root of both sides to solve for x.

The Square Root Property

For any positive number k,

if         

          x2 = k

then        

          x   =       or     x  =  -

 


Example

Solve

        x2 - 6  =  0

 

Solution

First add 6 to both sides

        x2  =   6  

Next use the square root property

        x  =          or        x  =  -

Example

Solve

        (x - 3)2 + 5  =  12

Solution

        (x - 3)2   =  7                               Subtract 5 from both sides

        x - 3  =     or    x - 3  =  -    Use the square root property

        x  =  3 +   or    x  =  3 -      Add 3 to both sides

 

Caution:  The square root property cannot be directly applied in a quadratic that has a middle term such as

                        x2 + 5x - 2

Completing The Square

We have seen that the square root property only worked when the middle term was zero.  For example if

        3(x - 1)2 - 3  =  0 

then we can use the square root property.  A quadratic is said to be in standard form if it has the form

        a(x - h)2 + k        Standard Form of a Quadratic

 

If we are given a quadratic in the form

        ax2 + bx + c

We would like to put the quadratic into standard form so that we can use the square root property.  We call the process of putting a quadratic into standard form Completing the Square.

 

Below is a step by step process of completing the square.

 

Example  

Complete the Square

        2x2 - 8x + 2 = 0

 

Solution

  1. Factor the leading coefficient from the first two terms:  

    2(x2 - 4x) + 2
  2. Calculate b/2:  

         -4
                 =  -2                b is the coefficient in front of the "x" term.
             2
  3. Square the solution above:  

    22  = 4 
  4. Add and subtract answer from part three (the magic number) inside parentheses: 

            2(x2 - 4x + 4 - 4) + 2
  5. Regroup:  

    2[(x2 - 4x + 4 ) - 4] + 2
  6. Factor the inner parentheses using part two as a hint:  

    2[(x - 2)2 - 4] + 2
  7. Multiply out the outer constant:    

    2(x - 2)2 - 8 + 2
  8. Combine the last two constants:  

    2(x - 2)2 - 6
  9. Breath a sigh of relief.

Example

Complete the square

        3x2 + 5x + 1

Solution

  1. 3(x2 + 5/3 x) + 1         Pulling a 3 out of a five is the same as dividing 5 by 3


  2. b/2  =  5/6


  3. (5/6)2  =  25/36         Square b/2


  4. 3(x2 + 5/3 x + 25/36 - 25/36) + 1    Add and subtract the magic number (b/2)2


  5. 3[(x2 + 5/3 x + 25/36) - 25/36] + 1    Regroup


  6. 3[(x + 5/6)2 - 25/36] + 1    Factor the first three terms


  7. 3(x + 5/6)2 - 25/12 + 1    Multiply the 3 through


  8. 3(x + 5/6)2 - 13/12        Note:  -25/12 + 1 = -25/12 +12/12 = -13/12

 

Exercises:  

Complete the square 

  1. 3x2 - 12x + 6         (x - 6)^2 -30
  2. 2x2 - 2x + 4             2(x - 1/2)^2 +7/2
  3. 4x2 + 4x - 3           4(x + 1/2)^2 - 4

 

Practice completing the square

 

Completing the Square to Solve a Quadratic Equation



Example

Solve

        x2 + 2x - 5  =  0
Solution

        We see that there is a middle term, 2x, so the square root property will not work.  We first complete the square.  We have 

        (b/2)2  =  1

        x2 + 2x + 1 - 1 - 5  =  0                   Adding and subtracting 1

        (x + 1)2 - 6 = 0                                Factoring the first three terms
Now we can use the square root property

        (x + 1)2  =  6                                    Adding 6 to both sides

        x + 1  =      or    x + 1  =  -   Taking the square root of both sides

        x  =  -1 +     or    x  =  -1 -    Subtracting 1 from both sides

 


Example

Solve

        x2 + 6x + 13  =  0



Solution

        We see that there is a middle term, 6x, so the square root property will not work.  We first complete the square.  We have 

        (b/2)2  =  9

        x2 + 6x + 9 - 9 + 13  =  0        Adding and subtracting the magic number 9

        (x + 3)2 + 4 = 0        Factoring the first three terms
Now we can use the square root property

        (x + 3)2  =  -4         Subtract 4 from both sides

        x + 3  =     or    x + 3  = -    Taking the square root of both sides

        x  =  -3 +     or    x  =  -3 -    Subtracting 1 from both sides

Notice that is not a real number but we can still write the imaginary solutions since

        =  2i

The final solutions are

        x  =  -3 + 2i    or    x  =  -3 - 2i

 

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