Completing The Square

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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,

x² = k

then

x = \(\sqrt{k}\) or x = -\(\sqrt{k}\)

Example

Solve

x² - 6 = 0

Solution

First add 6 to both sides

x² = 6

Next use the square root property

x = \(\sqrt{6}\) or x = -\(\sqrt{6}\)

Example

Solve

(x - 3)² + 5 = 12

Solution

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

x - 3 = \(\sqrt{7}\) or x - 3 = -\(\sqrt{7}\) Use the square root property

x = 3 + \(\sqrt{7}\) or x = 3 - \(\sqrt{7}\) 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

Factor the leading coefficient from the first two terms:

2(x²- 4x) + 2
Calculate b/2:

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

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

2(x²- 4x + 4 - 4) + 2
Regroup:

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

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

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

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

Example

Complete the square

3x² + 5x + 1

Solution

3(x² + 5/3 x) + 1 Pulling a 3 out of a five is the same as dividing 5 by 3
b/2 = 5/6
(5/6)² = 25/36 Square b/2
3(x² + 5/3 x + 25/36 - 25/36) + 1 Add and subtract the magic number (b/2)²
3[(x² + 5/3 x + 25/36) - 25/36] + 1 Regroup
3[(x + 5/6)² - 25/36] + 1 Factor the first three terms
3(x + 5/6)² - 25/12 + 1 Multiply the 3 through
3(x + 5/6)² - 13/12 Note: -25/12 + 1 = -25/12 +12/12 = -13/12

Exercises:

Complete the square

3x² - 12x + 6
2x² - 2x + 4
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 =  \(\sqrt{6}\)    or    x + 1 = -\(\sqrt{6}\)   Taking the square root of both sides

x = -1 + \(\sqrt{6}\) or x = -1 - \(\sqrt{6}\) 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 = \(\sqrt{-4}\)     or    x + 3 = - \(\sqrt{-4}\)    Taking the square root of both sides

x = -3 + \(\sqrt{-4}\) or x = -3 - \(\sqrt{-4}\) Subtracting 1 from both sides

Notice that \(\sqrt{-4}\)is not a real number but we can still write the imaginary solutions since

\(\sqrt{-4}\) = 2i

The final solutions are

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

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