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7.3 Completing the Square

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2 2

(square of a (perfect binomial) square

trinomia

x+2 = x 4

)

4x

l

1. 2 2

(square of a (perfect binomial) square

x 6 = x 12x 36

trinomial)

2.

In this section we will learn another method called completing the square. This method will give us the power to solve any quadratic equation. The trick to this section is getting the binomial squared on the LHS. Let’s observe the square of a binomial on the LHS.

So far we have solved some quadratic equations by factoring and by the Square Root Property (i.e. taking the square root both of sides). However, not all quadratic equations are factorable so we can not always depend on factoring.

22Examples: x 56 , x 3 24

Also, to use the Square Root Property, we needed to have the x squared on the LHS, or a binomial squared on the LHS.

Note that on the RHS of the trinomial, the last term is ½ of the coefficient of the middle term squared.

214 2, then 2 4

2 21

( 12) 6, then ( 6) 362

,

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1. Move the constant to the RHS, leave a space after the bx term.

+ 9 +9

216 3, -3 9

2

2x 6x 2

x 3 x 3 11

2x 3 11

x 3 11 Answer: 3 11

Solving Quadratic Equations by Completing the Square (C.T.S.)

Procedure: To solve a quadratic equation of the form ax2+bx+c=0 by C.T.S.

2Example 1. Solv x 6x 2e: 0

2. Divide by ‘a’ on both sides to get x2 on the LHS. In this example, it is not necessary.3. Multiply ‘b’ (the the coefficient of the x-term) by ½, then square the result. Note: This number will always be positive since any real number squared is positive. Add this number to both sides.4. The LHS is a perfect square trinomial. It can be factored and written as the square of a binomial. Also simplify the RHS.5. Then, solve by using the Square Root Property.

2Solve completing the square (C.T.S.): x 4x 10 0

Your Turn Problem #1

Answer: 2 14

3



+ 25 +25 21

10 5, -5 252

Solution:

2Example 2. Sol x 10: xe 1v 0

2x 10x 1

Move the constant to the RHS, take ½ of the middle number and add it to both sides.

Factor the LHS and write it as a binomial squared. Simplify the RHS.

x 5 x 5 24

2x 5 24 Now use the square root property.

x 5 2 6

Answer: 5 2 6

2Solve by Completing the Square: x 12x 4 0Your Turn Problem #2

Answer: 6 2 10

4



+ 1 +1

212 1, 1 1

2

Solution:

2Example 3. Sol x 2x 0v 3e: 1

2x 2x 13

Move the constant to the RHS, take ½ of the middle number and square it. Add this result to both sides.


x 1 x 1 12

2x 1 12

Now use the square root property.

x 1 12

Answer: 1 2i 3

x 1 i 12

x 1 2i 3

x 1 2i 3

2Solve by Completing the Square: x 4x 9 0


Answer: 2 i 5

5



2Example 4. So xl e 3v : 5x 0

Solution:

2

1 5 5 255 ,

2 2 2 4

5 37x

2 4

2x 5x 3

Move the constant to the RHS, take ½ of the middle number and square it. Add this result to both sides. Fractions are fine. Leave improper and don’t convert to decimal.


5 5 3 25x x

2 2 1 4


254

254

25 12 25

x2 4 4

25 37

x2 4

5 37x

2 2

5 37Answer:

2

5 37x

2 2

Since it is not a complex number, the solution is written as a single fraction.

2Solve by Completing the Square: x 7x 1 0


7 53Answer:

2

6



2Example 5. Sol 3x: 4ve 2x 0

Solution:

2

1 2 1 1 1,

2 3 3 3 9

23x 2x 4 Move the constant to the RHS. Then, divide by 3 to get the x2 on the LHS.



19

19

1 1 12 1x x

3 3 9 9

21 13

x3 9

1 13x

3 9

1 13x

3 3

1 13Answer:

3

1 13x

3 3

Since it is not a complex number, the solution is written as a single fraction.

Take ½ of the middle number and square it. Add this result to both sides

3 3 3

2 2 4x x

3 3

2Solve by Completing the Square: 3x 2x 2 0


1 7Answer:

3

The End.5-30-07

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