(Effective Alternative Secondary Education)

MATHEMATICS II

MODULE 2Quadratic Equations

BUREAU OF SECONDARY EDUCATIONDepartment of Education

DepEd Complex, Meralco Avenue, Pasig City

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Module 2Quadratic Equation s

What this module is about This module is about quadratic equations. This is a continuation of the

previous module. In this module you will learn two more methods of solving quadratic equations. You have to be proficient in the square root method of solving quadratic equations in order to understand the two methods to be discussed. As you go over the exercises, you will develop skills in solving quadratic equations and eventually use these skills in solving problems.

What you are expected to learnThis module is designed for you to:

1. Solve quadratic equations by completing the square.2. Derive the quadratic formula.3. Solve quadratic equations using the quadratic formula.

How much do you knowA. Solve the following by completing the square:

1. x² - 8x - 15 = 02. x² - 2x - 7 = 03. x² + 3x = 1

4. x² + x = 0

5. 6t² - 5t = t² + 3t – 1

B. Solve the following by the quadratic formula.

1. x² - x – 12 = 02. x² -4x = 03. 3x² + 2 = 04. 3x² - x = 15. x² = 1 – 2x

2

What you will do

Lesson 1

Completing the Square

Completing the square is a method used to make a quadratic expression a perfect square trinomial. This method is used to solve quadratic equations which cannot be solved by factoring.

The method of completing the square is based on the special product:

x2 + 2px + p2 = (x + p)2

x2 + 2px + p2 is a perfect square trinomial because it is the product of (x + p)2 , the square of a binomial. In this trinomial, notice that if you take one-half the coefficient of the x in the middle term and square it, you get the third term.

x2 + 2bx + b2

21 (2b)2 = (b)2 = b2

Therefore, to make an expression such as x2 + 12x into a perfect square trinomial, you must take one-half of 12, square it and add it to x2 + 12x.

If you follow the procedure, here is how it is done.

x2 + 12x

x2 + 12x + ________ (12)2 = (6)2 = 36

x2 + 12x + 36

Hence, x² + 12x + 36 is now a perfect square trinomial which when factored as (x + 6)(x + 6) can also be written as (x + 6)2.

Let’s look at some more examples:

Examples:

Complete the square of the following expressions:

a) x2 + 4x b) x2 - 6x c) x2 - 5x d) 2x2 - 3x

a) x2 + 4x

3

Solution:x2 + 4x + ______ The coefficient of the x term is 4.

x2 + 4x + (4)2 Take of 4 or divide 4 by 2.

x2 + 4x + (2)2 Square 2.

x2 + 4x + 4

Note that x2 + 4x + 4 = (x + 2)2

b) x2 - 6x Solution:

x2 - 6x + ______ The coefficient of the x term is -6.

x2 - 6x + (-6)2 of -6 is -3

x2 –6x + (-3)2 Square -3

x2 - 6x + 9

Note that x² - 6x + 9 = (x - 3)2

c) x2 - 5x

Solution: x2 - 5x + _______ The coefficient of the x term is -5

x2 - 5x + (-5)2 of -5 is

x2 –5x + ( )2 Square 25

x2 - 5x +

Note that x2 - 5x + = (x - )2

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) 2x2 - 3x

4

Solution: 2x2 - 3x

x2 – x + _____

x2 – x + (-23

)2

x2 – x + -43

2 Square -

x2 – x +

Note that x2 – x + = (x - )2

Try this out

Find the third term to be added to make each expression a perfect square trinomial. Write the result as a square of a binomial.

A. 1. x2 + 4x2. x2 + 6x3. x2 - 10x4. x2 - 8x5. x2 + 12x

B. 1. x2 + x2. a2 – 3a3. b2 - 13b

4. b² + b

5. c² - c

C. 1. 2x2 + 4x2. 3x2 - 6x3. 3x2 - x4. 4x2 - 3x5. 5x2 + 2x

5

Divide the terms by 2 to make the leading coefficient 1.

Complete the square:

of - is -

Lesson 2

Solution of Quadratic Equations by Completing the Square

You are now ready to solve equations by completing the square. In this method, the left-hand side of a quadratic equation is solved by completing the square so that it becomes a perfect square trinomial which can be written in the form (x + p)2 = d.

Example 1: Solve by completing the square:

x2 + 8x – 5 = 0

Solution: Note that the left member of the equation x2 + 8x – 5 = 0 is not a perfect trinomial square. Separate -5 from the left member by adding +5 to both sides of the equation.

x2 + 8x – 5 = 0

You now have x² + 8x on the left-hand side of the equation. Now, make the left-hand side a perfect square trinomial. You can do this by adding the square of one-half of the coefficient of x. So that,

x2 + 8x + (8)2 . of 8 is 4.

x2 + 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.

x2 + 8x = 5x2 + 8x + (4)2 = 5 + (4)2

x2 + 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

6

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.

x+4 = Taking the square root

x = -4 Subtracting 4 on both

sides

x = -4 + or x = -4 -

The solutions are -4 + and -4 - .

Check::If x = -4 +

x² + 8x – 5 = 0 0 = x² + 8x - 5

= (-4 + )² + 8(-4 + ) – 5= 16 - 8 + ( )² - 32 + 8 – 5= 16 + 21 – 32 – 5= 37 – 37

= 0If x = -4 -

x² + 8x – 5 = 0 0 = x² + 8x - 5

= (-4 - )² + 8(-4 - ) – 5= 16 + 8 + ( )² - 32 - 8 – 5= 16 + 21 – 32 – 5= 37 – 37

= 0The solutions check.

Example 2: Solve by completing the square.

4x2 + 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.

4x2 + 4x - 3 = 0

7

x2 + x - = 0 Divide both sides by 4

x2 + x = 43

Add to both sides

x2 + x +21

(1)2 = 43

+21

(1)2

x2 +x +41

= 43

+ 41

(x + 21

)2 = 1 Factor

x + 21

= Use square root method

x + 21

= 1

x = -21

1

x = -21

+ 1 x = -21

- 1

x = 21

x = -

Note that this equation can be solved by factoring:

4x2 + 4x - 3 = 0

(2x – 1)(2x + 3) = 0

2x – 1 = 0 or 2x + 3 = 02x = 1 2x = -3

x = 21

x = -23

Check:

If x = 4x2 + 4x - 3 = 0

4( )² + 4(21

) – 3 = 0

4( ) + 24

– 3 = 0

8

Add (1)2 to both sides

1 + 2 – 3 = 03 – 3 = 0 0 = 0

If x = - 4x2 + 4x - 3 = 0

4(-23

)² + 4(-23

) – 3 = 0

4( ) – – 3 = 0

9 – 6 – 3 = 0 3 – 3 = 0

Both solutions check.

Example 3:

Solve by completing the square:

2x2 - 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:

2x2 - 5x –3 = 0

x2 – x – = 0 Divide both sides by 2

x2 – x = Add 3/2 to both sides

x2 – x + (- )2 = + (- )2

x2 – x + - 2 = + - 2

x2 – x + = + Simplify

(x – )2 = +

(x – )2 = Adding the fractions

9

Add (- )2 to both sides

Factor the left-hand side & get the LCD of the right-hand side of the equation.

(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 – .

Note that this equation can also be solved by factoring:

2x² - 5x – 3 = 0

( 2x + 1 )( x – 3 ) = 0

2x + 1 = 0 or x – 3 = 0

2x = -1 x = 3

x = -

Check:

If x = - 2x² - 5x – 3 = 0

2(- )² - 5(- ) – 3 = 0

2( ) + –3 = 0

+ – = 0

– = 0

0 = 0

If x = 3 2x² - 5x – 3 = 0

2(3)² - 5(3) – 3 = 0

10

2(9) + 15 – 3 = 0

18 + 15 – 3 = 0

3 – 3 = 0

0 = 0

The solutions check.

Example 4:

Use completing the square to solve 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)2 = 3 + (2)2

x² + 2x + 1 = 3 + 1 (x + 1)² = 4 x + 1 = Solve for x x = -1 2

x = -1 + 2 or x = -1 –2x = 1 or x = -3

The solutions are 1 and -3.

Note that the equation can be solved by factoring.

-x² -2x + 3 = 0-(x² + 2x – 3) = 0 x² + 2x – 3 = 0(x + 3)(x – 1) = 0

x + 3 = 0 or x – 1 = 0 x = -3 or x = 1

Check::If x = -3 -x² -2x + 3 = 0

11

-(-3)² - 2(-3) + 3 = 0

-(9) + 6 +3 = 0

-9 + 9 = 0

0 = 0

If x = 1 -x² -2x + 3 = 0

-(1)² - 2(1) + 3 = 0

-1 – 2 +3 = 0

-3 + 3 = 0

0 = 0The solutions check.

Example 5:

Use completing the square to solve equation:

-2x2 - 4x +2 = 0.

Solution: This equation cannot be solved by factoring.

-2x2 -4x +2 = 0

x2 +2x -1 = 0 Divide both sides by -2

x2 + 2x = 0 Add 1 to both sides

x2 + 2x+ (2)2 = 1 + (2)2

x2 + 2x +1 = 1+ 1

(x + 1 )2 = 2 Factor and Simplify

x + 1 =

x = -1 Solve for x

x = -1+ and x = -1 -

The solutions are -1+ and -1+ .

Check: If x = -1+ -2x2 - 4x + 2 = 0

-2(-1+ )² - 4(-1+ ) + 2 = 0

12

Add (2)2 to both sides

to complete the square

Solve by the square root method

-21 - 2 + ( )² + 4 - 4 + 2 = 0 -2 + 4 - 2(2) + 4 - 4 + 2 = 0 -2 - 4 + 4 + 2 = 0

-6 + 6 = 0 0 = 0

If x = -1 - -2x2 - 4x + 2 = 0

-2(-1 - )² - 4(-1 - ) + 2 = 0 -21 + 2 + ( )² + 4 + 4 + 2 = 0

-2 - 4 - 2(2) + 4 + 4 + 2 = 0 -2 – 4 + 4 + 2 = 0

-6 + 6 = 0 0 = 0

Both solutions check.

From the preceding examples, you must be able to identify the steps in solving quadratic equations by completing the square method.

To solve an equation by completing the square, follow these steps:

1. Make sure that coefficient of x2 is 1. If it is not, make it 1 by dividing both sides of the equation by the coefficient of x2.

2. Isolate the constant (numerical term) by transposing it on the right-hand side of the equation.

3. Complete the square:a) Identify the coefficient of x.b) Take half the coefficient of x.c) Square half the coefficient of xd) Add the numbers in c to both sides of the equation

4. Factor the left side and simplify the right sides of the equation.

5. Solve the resulting equation by square root method.

6. Check the solution.Try this out

Solve the following by completing the square.

For Nos. 1 – 15, see Example 1

13

1. x2 + 8x + 6 = 0

2. x2 – 4x – 3 = 0

3. x2 – 10x = 15

4. u2 – 5u – 2 = 0

5. x2 – 6x = 19

6. a2 – 4a – 5 = 0

7. x2 + 10x + 20 = 0

8. x² -2x – 5 = 0

9. a² - 8a – 20 = 0

10. z² + 5 = 7z

11. x² + 8x – 9 = 0

12. x² + 6x – 7 = 0

13. x² + 6x + 8 = 0

14. x² + 5x – 6 = 0

15. x² = 4x + 3

For Nos. 16 – 20, see Examples 2 and 3

16. 2x2 = a – 4x

17. 3x2 = 12 – 6x

18. 3x² = 2 – 5x

19. 2x² = 5x + 3

20. 4x² = 2 – 7x

For Nos. 21 – 25, see Examples 4 and 5

21. –x² - 12x = 6

22. -2z² + 12z – 4 = 0

23. 2x(x + 3) = 8

24. x(x + 3) -1/2 = -2

25. 20x² - 11x – 3 = 0

Lesson 3

14

The Quadratic Formula

In this lesson, you will see how completing the square will be used to derive the Quadratic Formula. This method works easily in solving any quadratic equation.

To do this, let us begin with solving a Quadratic Equation of the form ax 2 + bx + c = 0 by completing the square.

Example 1 Solve 2x2 + 7x + 4 = 0 by completing the square.

Solution: 2x2 + 7x + 4 = 0

x2 + x + 2 = 0 Divide both sides of the eq’n by 2

x2 + x = - 2 Add -2 to both sides of the eq’n

x2 + x + ( )2 = -2 + ( )2

x2 + x + = -2 +

(x + )2 = - +

(x + )2 =

x + =

x + = Solve for x.

x = -

x = and x =

Now, study the following derivation of the quadratic formula by the method of completing the square. The steps is just like what has been done in Example 1 so that you can see that the process is the same. The only difference is that you will be working with letters instead of numbers.

15

Add the square of of .

Factor the left-hand side and simplify the right –hand side by finding the LCD, which is 16.

By the square root method

ax2 +bx + c = 0

x2 + x + = 0 Divide both sides by a.

x2 + x = - Subtract from both sides.

x2 + x + ( )2 = - + ( )2 Complete the square.

x2 + x + = - +

(x + )2 = –

x + = (5)

x + = Solve for x.

x = -

or x =

Presto! What had been derived is the Quadratic Formula.

The quadratic formula tells you that if you have a quadratic equation in standard form ax2 + bx + c =0, then all you have to do is substitute the values of a, b and c into the formula to get the solutions.

Example 2

Solve by using the quadratic formula

16

Factor the left-hand side and simplify the right-hand side by finding to LCD which is 4a.

Add the square of of .

Solve the equation by the square root method

The Quadratic Formula:

x =

x2 - 3x - 5 =0

Solution: The equation is already in standard formula. To use the formula you must first identify a, b, and c.

a = 1 b = -3 c = -5

Substituting these values into the formula, you should get

x =

= )1(2

)5)(1(4)3()3( 2

=

=

The solution are and

Check:

If x = x2 - 3x - 5 = 0

( )² - 3( ) - 5 = 0

- 2

2939 – 5 = 0

- – 5 = 0

- 5 = 0

- 5 = 0

5 - 5 = 0 0 = 0

If x = x2 - 3x - 5 = 0

17

( )² - 3( ) - 5 = 0

- 2

2939 – 5 = 0

- – 5 = 0

- 5 = 0

- 5 = 0

5 - 5 = 0 0 = 0

You may find that checking is a little bit tedious but its fun doing since you get to apply your skills in simple arithmetic.

Notice that the quadratic formula is a lot easier to use than completing the square. This is because the formula contains all the steps necessary if you were using completing the square. Here are some things to watch out when using the Quadratic Formula:

1. If b is positive, then the – b that appear in the formula will be negative. If b is negative, then the value of – b in the formula will be positive.

2. The quantity 2a in the formula is the denominator of the entire expression -b .

Example 3:

Solve by using the quadratic formula

3x2 – 3x =5

Solution: Begin by writing the equation into the standard form

3x2 +11x = 4

3x2 + 11x - 4 = 0

Substitute the value of a, b and c into the Quadratic Formula

18

a = 3 b = 11 c = -4

x = a

acbb2

42

Then we have,

x =

=

=

x =

x = and x =

x = = = = -4

The solution is and -4.

The checking is left for you. Example 4

Solve by the quadratic formula:

3 – 2t = t2

Solution: Arrange the equation in standard form and identify a, b, and c.

t2 + 2t - 3 = 0

a = 1 b = 2 c = -3

Substitute in the quadratic formula and solve for t:

t = a

acbb2

42

19

t =

t =

=

t = 2

42

t = and t =

= =

t = 1 t = -3

The solution are 1 and -3.

This equation can be solved by factoring. Try solving and check..

Example 5

Solve for x : 2x2 – 5x + 7 = x(2x - 3)

Solution: Be careful! This does not mean that just because there is an x2

in the equation that the equation 2x2 – 5x + 7 = x(2x - 3) must be a quadratic equation.

Putting the equation in standard form, then

2x2 – 5x + 7 = x(2x- 3)

2x2 – 5x + 7 = 2x2 - 3x - 2x2 + 3x -2x2 + 3x by addition

-2x + 7 = 0 -2x = -7

x =

As you can see, it’s not a quadratic equation at all. Example 5 clearly shows, the method of solving an equation depends on the type of equation you are dealing with. Look carefully at the equation before deciding the method to apply.

20

Example 6

Solve by the quadratic formula:

x2 + 3x+ 4 =0

Solution: The equation is already in standard form, so that

a = 1 b = 3 c = 4

x =

x =

=

x = 2

73 Not real solution

As soon as you see that the answer involves the square root of a negative number, you can stop and say that the equation has no real solution.

Try this out

A. Write each equation in standard form, if necessary. Then, determine the values of a, b and c. Do not solve the equation.

1. x² + 4x + 3 = 02. x² - x – 4 = 03. 3x² - 2x + 7 =04. 4x² + 7x – 3 = 05. 4y² = 2y – 16. 2x = 3x² + 47. x(3x – 5) = 28. y(5y + 10) = 8

Solve the equations using the Quadratic Formula.

B. 1. x2 + 3x - 5 =02. x2 + 5x - 2 =03. y2 + 4y - 6 =0

21

4. y2 + 2y - 5 =05. u2 – 2u + 3 =06. u2 - 3u + 3 =07. t2 - 7t = 68. t2 + 6 = 6t

C.1. 5x2 - x = 22. 7x2 - 3 = x3. 3x2 + 5x + 2 =04. 2x2 - 3x - 1 =05. t2 - 3t + 4 = 2t2 + 4t - 36. (5w + 2) (w - 1) = 3w + 17. (x + 2)2 = x(x + 4)

D. Find out what is wrong with the following solution:

x² - 3x – 1 = 0

x =

x =

=

Let’s Summarize

1. Most types of quadratic equations can be solved by completing the square.

2. In completing the square, it is important to remember that

a. the coefficient of x2 should be 1; if not, divide both sides of the equation by the coefficient of x2.

b. the term to be added is the square of one-half the coefficient of x in the equation.

3. The quadratic formula is

x =

4. Practically, any quadratic equation can be solved by the quadratic formula.

22

What have you learned A. Solve by completing the square:

1. x² - 4x – 2 = 02. x² - 2x – 4 = 03. x² + 2x = 04. 6(x² - 1) = 5x5. (x + 2)2 – 4x = 8

B. Solve using the quadratic formula:

1. x² + 6x + 9 = 02. x² -4x = 03. x² - 7 = 04. x² + 8x + 12 = 05. 2x² + x = 5

23

Answer Key

How much do you know

A. 1. 4 ±

2. 1 ± or 1 ± 2

3.

4. 0, -

5.

B. 1. 4, -3

2. 0, 4

3. no real solution

4.

5. -1 ± 2

Try this out

Lesson 1

A. 1. 4, (x + 2)2

2. 9, (x + 3)2

3. 25, (x - 5)2

4. 16, (x - 4)2

5. 36, (x + 6)2

B. 1. 41

, (x + )2

2. , (a - )2

3. , (b - )2

24

4. , (b + )2

5. , (c - )2

C. 1. 1, (x + 1)2

2. 1, (x - 1)2

3. , (x - )2

4. , (x - )2

5. , (x - 251

)2

Lesson 2B. 1. – 4 ±

2. 2 ± 7

3. 5 ± or 5 ± 2

4.

5. 3 ± or 3 ± 2

6. 5, -1

7. -5 ±

8. 1 ±

9. 10, -2

10.

11. -9, 1

12. -7, 1

13. -2, -4

14. 1, -6

15. 2 ± 7

16. -1±

17. -1 ± /5

25

18. -2,

19. 3, -

20. -2,

21. -6 ±

22. 3 ±

23. -4, 1

24.

25. ¾, 1/5

Lesson 3

A.1. SF a = 1 b = 4 c = 3

2. SF a = 1 b = -1 c = -4

3. SF a = 3 b = -2 c = 7

4. SF a = 4 b = 7 c = -3

5. 4y2 -2y + 1=0 a = 4 b = -2 c = 1

6. 3x2-2x +4 =0 a = 3 b = -2 c = 4

7. 3x2 -5x -2 = 0 a = 3 b = -5 c = -2

8. 5y2 + 18y- 8 =0 a = 5 b = 10 c = -8

B.

1.

2.

3.

4. -1 ±

5. no real solution

6. no real solution

7.

26

8. 3 ±

C.

1.

2.

3. no real solution

4.

5.

6.

7. no real solution

D. The solution should be:

x² - 3x – 1 = 0

x =

x =

=

What have you learned

A. 1. 2 ±

2. 1 ±

3. -2, 0

4. , -

5. ±2

B. 1. -3

27

2. 4, 0

3. ±

4. -2, -6

5.

28