Solving Quadratic Equations by Completing the Square, Exercises of Mathematics

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Mathematics

First Quarter

Module 2C: Solving Quadratic

Equations by Completing the

Square

Module 2 C

Find My X

What This Module Is All About

This learning material deals with solving quadratic equations by completing the

square. As you go through this lesson your skill in finding the solutions of a quadratic

equation by completing the square will be developed.

What You Are Expected To Learn

After going through this module, the learners should be able to solve quadratic

equations by: (c) completing the square (M9AL-Ia-b- 1 )

How Much Do You Know (Pre-test)

A. Directions: Choose the letter that corresponds to the correct answer.

1. Which of the following can be the third term in 𝑥

2

• 10 𝑥+___ to make it a perfect

square trinomial?

A. 25 C. 10

B. 20 D. 5

2. Which of the following square of a binomial is equal to 𝑥

2

A. (𝑥 − 6 )(𝑥 + 6 ) = 0 C. (𝑥 + 6 )

2

B. (𝑥 + 6 )(𝑥 − 6 ) = 0 D. (𝑥 − 6 )

2

3. Which of the following is the next step in solving quadratic equation 𝑥

2

• 10 𝑥 = 11 by

completing the square?

A. Add – 11 to both sides

B. divide the numerical coeffient of the linear term by 2 or multiply it by

1

2

C. square the quotient of 10 and 2

D. add + 25 to both sides

4. Find the solutions of quadratic equation 𝑥

2

• 10 𝑥 + 22 = 0 by completing the square.

A. − 5 −

3 C. + 5 −

B. + 5 − √ 3 , − 5 − √ 3 D. + 5 − √ 3 , + 5 + √ 3

B. Determine a number that must be added to make each of the following a perfect

square trinomial:

2

+ 2 𝑥 + _______

2

+ 20 𝑡 + _______

Hence, 𝑥

2

• 12 𝑥 + 36 is now a perfect square trinomial which when factored as

(𝑥 + 6 )(𝑥 + 6 ) and can also be written as (𝑥 + 6 )

2

Let’s look at some more examples:

Complete the square of the following expressions:

a) x

2

• 4x b) x

2

• 6x c) x

2

• 5x d) 2x

2

• 3x

a) x

2

• 4x

Solution:

x

2

• 4x + ______ The coefficient of the x term is 4.

x

2

• 4x + 

2

1

2

Take

2

1

of 4 or divide 4 by 2.

x

2

• 4x + (2)

2

Square 2.

x

2

• 4x + 4

Note that x

2

• 4x + 4 = (x + 2)

2

b) x

2

• 6x

Solution:

x

2

• 6x + ______ The coefficient of the x term is - 6.

x

2

• 6x +

2

1

2

2

1

of - 6 is - 3

x

2

• 6x + (-3)

2

Square - 3

x

2

• 6x + 9

Note that x² - 6x + 9 = (x - 3)

2

c) x

2

• 5x

Solution:

x

2

• 5x + _______ The coefficient of the x term is - 5

x

2

• 5x + 

2

1

2

2

1

of - 5 is

2

− 5

x

2

• 5x + (

2

− 5

2

Square

x

2

• 5x +

Note that x

2

• 5x +

= (x -

2

5

2

2

− 5

You will notice that the leading coefficient in the first three examples is 1. So, before

proceeding to completing the square, transform the leading coefficient to 1. The leading

coefficient of the next example is not equal to 1.

d) 2x

2

• 3x

Solution:

2x

2

• 3x

x

2

x + _____

x

2

x + 

2

x

2

x + - 

2

Square -

x

2

x +

Note that x

2

x +

= (x -

2

Activity 1.

Directions: Determine a number that must be added to make each of the following a

perfect square trinomial.

2

+ 2 𝑥 + ____ 6. 𝑥

2

+ 11 𝑥 + ____

2

+ 20 𝑡 + _____ 7. 𝑥

2

− 15 𝑥 + ____

2

− 16 𝑟 + _____ 8. 𝑤

2

+ 21 𝑤 + ____

2

+ 24 𝑟 + _____ 9. 𝑠

2

2

3

𝑠 + _____

2

− 30 𝑥 + _____ 10. ℎ

2

3

4

ℎ + _____

Activity 2.

Directions: Express the following perfect square trinomials as a square of a binomial.

2

2

2

2

2

3

1

9

2

2

49

4

2

2

3

4

9

64

2

2

25

4

This time you’re now ready to solve quadratic equations by completing the square.

Divide the terms by 2 to make

the leading coefficient 1.

Complete the square:

of -

is -

Check the solutions obtained against the original equation 2 𝑥

2

Both the values of x satisfy the given equation. So, the solutions we had are

correct and valid.

Answer: The equation 2 𝑥

2

• 8 𝑥 − 10 = 0 has two solutions: 𝒙 = 𝟏 and 𝒙 = −𝟓.

Example 2 : Solve by completing the square:

2

Solution: Note that the left member of the equation 𝑥

2

• 8 𝑥 − 5 = 0 is not a perfect square

trinomial. Separate - 5 from the left member by adding +5 to both sides of the

equation.

2

You now have x² + 8x on the left side of the equation. Now, make the left side a

perfect square trinomial. You can do this by adding the square of one-half of the

coefficient of x. So that,

x

2

• 8x + 

2

1

2

2

1

of 8 is 4

x

2

• 8x + (4)

2

Square 4

Since you have done the first part in completing the square of the left member of

the equation, go back to the original equation.

x

2

• 8x = 5

x

2

• 8x + (4)

2

= 5 + (4)

2

x

2

• 8x + 16 = 5 + 16

(x + 4)

2

= 21

The equation is now of the form (x + p)

2

= d, and so the square root method may

then be used. Thus,

(x+4)

2

= 21

x+4 =  21 Taking the square root

x = - 4  21 Subtracting 4 on both sides

x = - 4 + 21 or x = - 4 - 21

For x = 1:

2

2

For x = - 5:

2

2

Since (4)

2

or 16 is added to the

left member, (4)

2

or 16 must also

be added to the right member by

Addition Property of Equality.

The solutions are - 4 + 21 and - 4 - 21.

Check:

If x = - 4 + 21

x² + 8x – 5 = 0

0 = x² + 8x - 5

0 = (-4 + 21 )² + 8(-4 + 21 ) – 5

0 = 16 - 8 21 + ( 21 )² - 32 + 8 21 – 5

If x = - 4 - 21

x² + 8x – 5 = 0

0 = x² + 8x - 5

0 = (- 4 - 21 )² + 8(- 4 - 21 ) – 5

0 = 16 + 8 21 + ( 21 )² - 32 - 8 21 – 5

The solutions are correct.

Example 3 : Solve by completing the square.

4x

2

• 4x - 3 = 0

Solution: Notice that the coefficient of x² is not 1. You need to make the coefficient

of x² equal to 1 first before you attempt to use completing the square.

Follow the steps on the right.

4x

2

• 4x - 3 = 0

x

2

• x -

= 0 Divide both sides by 4

x

2

• x = Add

4

3

to both sides

Example 4 :

Solve by completing the square:

2x

2

• 5x – 3 = 0

Solution: Again, just like Example 2, you need to make the coefficient of x² equal

to 1. Divide both sides of the equation by 2 to make the coefficient of

x² equal to 1. Then proceed as follows:

2x

2

• 5x – 3 = 0

x

2

x –

= 0 Divide both sides by 2

x

2

x =

Add 3/2 to both sides

x

2

x + 

2

2

x

2

x + -

2

=

2

x

2

x +

Simplify

(x –

2

=

(x –

2

Adding the fractions

(x –

2

= 

Using the square root method

x –

Add

to both sides.

x =

Solve for x

x =

and x =

x =

x = -

x = 3 x = -

The solutions are 3 and –

Check:

Add 

2

1

2

to both sides

Factor the left-hand side & get the

LCD of the right-hand side of the

equation.

If x = -

2x² - 5x – 3 = 0

If x = 3

2x² - 5x – 3 = 0

The solutions are correct.

Example 5 :

Use completing the square to solve the equation:

• x² - 2x + 3 = 0

Solution: In this equation, note that the coefficient of x² is – 1. This means that

you just cannot apply completing the square right away. You need to

make the coefficient of x² equal to 1. To do this, multiply both sides

of the equation by – 1. Thus,

• 1(-x² - 2x + 3) = 0(-1)

x² + 2x – 3 = 0

x ² + 2x + 

2

1

2

2

1

2

x² + 2x + 1 = 3 + 1

(x + 1)² = 4

x

2

• 2x +1 = 1+ 1

(x + 1 )

2

= 2 Factor and Simplify

2

=  2

x = - 1  2 Solve for x

x = - 1+ 2 and x = - 1 - 2

The solutions are - 1+ 2 and - 1+ 2.

Check:

If x = - 1+ 2

• 2x

2

• 4x + 2 = 0
• 2(-1+ 2 )² - 4(-1+ 2 ) + 2 = 0
• 2  1 - 2 2 + ( 2 )² + 4 - 4 2 + 2 = 0
• 2 + 4 2 - 2(2) + 4 - 4 2 + 2 = 0

If x = - 1 - 2

• 2x

2

• 4x + 2 = 0
• 2(- 1 - 2 )² - 4(- 1 - 2 ) + 2 = 0
• 2 1 + 2 2 + ( 2 )² + 4 + 4 2 + 2 = 0
• 2 - 4 2 - 2(2) + 4 + 4 2 + 2 = 0

Both solutions are check.

Solve by the square root

method

Try this out

Solve the following by completing the square.

1. x

2

• 8x + 6 = 0

2. x

2

• 4x – 3 = 0

3. x

2

• 10x = 15

4. u

2

• 5u +4 = 0

5. x

2

• 6x = 16

Activity 3

Find the solutions of each of the following quadratic equation by completing the

square:

2

2

2

2

2

2

2

2