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

2. 2x2 - 2x + 4

3. 4x2 + 4x - 3

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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