1 Topic 7.1.1 Solving Quadratic Equations by Factoring Solving Quadratic Equations by Factoring.

1

Topic 7.1.1Topic 7.1.1

Solving Quadratic

Equations by Factoring

Solving Quadratic

Equations by Factoring

2

Topic7.1.1

Solving Quadratic Equations by FactoringSolving Quadratic Equations by Factoring

California Standards:11.0 Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials.

14.0 Students solve a quadratic equation by factoring or completing the square.

What it means for you:You’ll solve quadratic equations by factoring.

Key words:• quadratic• factor• zero property

3

Topic7.1.1


In this Topic you’ll use all the factoring methods that you learned in Chapter 6 to solve quadratic equations.

4

Topic7.1.1

Quadratic Equations Have Degree 2


Quadratic equations contain a squared variable, but no higher powers — they have degree 2.

These are all quadratic equations, as the highest power of the variable is 2:

(i) x2 – 3x + 2 = 0 (ii) 4x2 + 12x – 320 = 0 (iii) y2 + 4y – 7 = 2y2 – 2y

5

Topic7.1.1


The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers, and a is not 0.

For example, in (i) above, a = 1, b = –3, and c = 2, while in (ii), a = 4, b = 12, and c = –320.

Example (iii) above is a quadratic in y, while the others are quadratics in x.

(i) x2 – 3x + 2 = 0

(ii) 4x2 + 12x – 320 = 0

(iii) y2 + 4y – 7 = 2y2 – 2y

6

Topic7.1.1

Guided Practice

Solution follows…


The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations.

a = –1, b = 5, c = –6

a = 6, b = 31, c = 35

1. –x2 + 5x – 6 = 0

2. 6x2 + 31x + 35 = 0

3. 4x2 – 12x + 9 = 0

4. 16x2 – 8x + 1 = 0

5. –x2 – 4x – 4 = 0

6. 64x2 + 48x + 9 = 0

a = 4, b = –12, c = 9

a = 16, b = –8, c = 1

a = –1, b = –4, c = –4

a = 64, b = 48, c = 9

7

Topic7.1.1

Guided Practice

Solution follows…


The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in these equations.

7. 6y2 + 28y + 20 = 5 – 6y2

8. 4x2 + 6x + 1 = 3x2 + 8x

9. 4(x2 – 5x) = –25

10. 3x(3x + 4) + 8 = 4

11. 7x(7x + 4) + 4x3 + 3 = 2(2x3 + 1)

a = 12, b = 28, c = 15

a = 1, b = –2, c = 1

a = 4, b = –20, c = 25

a = 9, b = 12, c = 4

a = 49, b = 28, c = 1

8

Topic7.1.1

Solving is Finding Values That Make the Equality True


An equation is a statement saying that two mathematical expressions are equal.

If an equation contains a variable (an unknown quantity), then solving the equation means finding possible values of the variable that make the equation a true statement.

For example, 7 + 2 = 9, 4x + 2 = 14, and x2 – 3x + 2 = 0 are all equations.

9

Topic7.1.1

Example 1

Solution follows…


Find a solution of the equation x2 – 3x + 2 = 0.

If you evaluate the above equation using, say, x = 3, then you get:

32 – (3 × 3) + 2 = 0

This is not a true statement (since the left-hand side equals 2).

So x = 3 is not a solution of the equation.

Solution

But if instead you substitute x = 1, then you get:12 – (3 × 1) + 2 = 0

This is a true statement.

So x = 1 is a solution of the equation x2 – 3x + 2 = 0.

10

Topic7.1.1

Guided Practice

Solution follows…


Determine which of the two values given is a solution of the equation.

x = 3

x = –2

5

31

21

4

3

8

12. –x2 + 5x – 6 = 0 for x = –3 and x = 3

13. 6x2 + 31x + 35 = 0 for x = – and x =

14. 4x2 – 12x + 9 = 0 for x = and x =

15. 16x2 – 8x + 1 = 0 for x = – and x =

16. –x2 – 4x – 4 = 0 for x = 2 and x = –2

17. 64x2 + 48x + 9 = 0 for x = – and x = 3

8

5

3

1

4

3

2

x = –5

3

x = 3

2

x = 1

4

x = –3

8

11

Topic7.1.1

Zero Property — if xy = 0, then x = 0 or y = 0 (or Both)


One way to solve a quadratic equation is to factor it and then make use of the following property of zero:

Zero Property

If the product mc = 0, then either:(i) m = 0,(ii) c = 0,(iii) both m = 0 and c = 0.

12

Topic7.1.1

Example 2

Solution follows…


Solve x2 + 2x – 15 = 0 by factoring.

Solution

x2 + 2x – 15 = (x – 3)(x + 5)

So either x = 3 or x = –5.

So if (x – 3)(x + 5) = 0, then by the zero property, either (x – 3) = 0 or (x + 5) = 0.

13

Topic7.1.1

Example

Solution follows…


Solve 2x2 + 3x – 20 = 0 by factoring.

Solution

3

2x2 + 3x – 20 = (2x – 5)(x + 4) = 0

So either x = or x = –4.5

2

14

Topic7.1.1

Guided Practice

Solution follows…


Solve each of these quadratic equations by using the zero property.

18. (2x + 7)(3x + 5) = 0

19. (x – 5)(x – 1) = 0

20. 49x2 – 1 = 0

21. 64a2 – 25 = 0

22. 4x2 + 8x + 3 = 0

23. 2x2 – 17x – 9 = 0

x = 5, 1

x = – , 91

2

x = – , –3

2

1

2

a = , –5

8

5

8

x = , –1

7

1

7

x = – , –7

2

5

3

15

Topic7.1.1

Guided Practice

Solution follows…


Solve each of these quadratic equations by using the zero property.

24. 4x2 – 11x – 3 = 0

25. 10x2 – x – 2 = 0

26. 2x2 + 11x + 12 = 0

27. 10x2 – 27x + 5 = 0

28. 3x2 – 17x – 28 = 0

29. 2x2 – x – 28 = 0

x = – , 31

4

x = , – 1

2

2

5

x = – , –43

2

x = , 5

2

1

5

x = – , 74

3

x = – , 47

2

16

Topic7.1.1

Using Factoring to Solve Quadratic Equations


1) First arrange the terms in the quadratic equation so that you have zero on one side.

3) Once done, you can use the zero property to find the solutions.

2) Then factor the nonzero expression (if possible).

17

Topic7.1.1

Example

Solution follows…


Solve x2 – 6x – 7 = 0.

Solution

4

The right-hand side of the equation is already zero, so you can just factor the left-hand side:

x2 – 6x – 7 = (x + 1)(x – 7)

So (x + 1)(x – 7) = 0.

Using the zero property, either x + 1 = 0 or x – 7 = 0.So either x = –1 or x = 7.

18

Topic7.1.1

Example

Solution follows…


Solve x2 + 2x – 11 = –3.

Solution

5

By adding 3 to both sides, the right-hand side becomes 0.

Now you can factor the left-hand side:

Since you have two expressions multiplied to give zero, you can use the zero property. That is, either x + 4 = 0 or x – 2 = 0.So either x = –4 or x = 2.

This time, you have to arrange the equation so you have zero on one side.

x2 + 2x – 11 = –3

x2 + 2x – 8 = 0

x2 + 2x – 8 = (x + 4)(x – 2) = 0

19

Topic7.1.1

Example

Solution follows…


Solve 3x2 + 168 = 45x.

Solution

6

Once again, the first thing to do is get zero on one side:3x2 + 168 = 45x

The left-hand side can be factored, which means you can rewrite this as: 3(x2 – 15x + 56) = 0,

or 3(x – 7)(x – 8) = 0

So using the zero property, either x – 7 = 0 or x – 8 = 0.So either x = 7 or x = 8.

3x2 – 45x + 168 = 0

20

Topic7.1.1

Guided Practice

Solution follows…


30. x2 – 2x – 15 = 0

31. x2 – 7x – 18 = 0

32. k2 + 10k + 24 = 0

33. 4m2 + 4m – 15 = 0

34. 8k2 – 14k = 49

35. 15k2 + 28k = –5

x = –3, 5

x = –2, 9

k = –6, –4

Solve each of these equations.

m = , –3

2

5

2

k = – , 7

4

7

2

k = – , –5

3

1

5

21

Topic7.1.1

Guided Practice

Solution follows…


36. 6y2 + 28y + 20 = 5 – 6y2

37. 4x2 + 6x + 1 = 3x2 + 8x

38. 4(x2 – 5x) = –25

39. 3x(3x + 4) + 8 = 4

40. x(x + 4) + 9 = 5

41. x(x – 5) + 3 = –3

x = 1

x = –2

x = 2, 3


y = – , –5

6

3

2

x =5

2

x = –

2

3

22

Topic7.1.1

Guided Practice

Solution follows…


42. 6x(3x – 4) – 7 = –15

43. 3 = 2 – 12x(3x – 1)

44. 7x(7x + 2) + 4x3 + 3 = 2(2x3 + 1)

45. (2x + 9)2(x + 3)(x + 1)–1(x + 3)–1(x + 1) = 0

46. 2x(3x + 3) + 4(x + 1) = 1 + 2x + 2x2


x =2

3

x =1

6

x = –

1

7

x = – 9

2

x = – , –1

2

3

2

23

Topic7.1.1

Independent Practice

Solution follows…


The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are numbers. Identify a, b, and c in the quadratic equations below.

1. 4x2 + 20x + 9 = 0

2. x2 – 9x + 8 = 0

3. 2x2 + 5x = 35 + 14x

4. x(2x + 3) = 5(2x + 3)

5. y(2y + 7) = 9(2y + 7)

6. (x + 2)(x – 2) = 3x

a = 4, b = 20, c = 9

a = 1, b = –3, c = –4

a = 2, b = –11, c = –63

a = 2, b = –7, c = –15

a = 2, b = –9, c = –35

a = 1, b = –9, c = 8

24

Topic7.1.1


Solution follows…


Use the zero product property to solve these equations.

7. (2y + 9)(2y – 3) = 0

8. (2a + 5)(a – 11) = 0

9. (y – 3)2(y – 5)(y – 3)–1 = 0

10. (y – 4)3(2y – 9)2(y – 4)–3(y – 7)(2y – 9)–1 = 0

y = 3, 5

y = – , 9

2

3

2

a = – , 11 5

2

y = , 79

2

25

Topic7.1.1


Solution follows…


Solve the following equations.

11. 4x2 + 20x + 9 = 0

12. x2 – 9x + 8 = 0

13. 2x2 + 5x = 35 + 14x

14. x(2x + 3) = 5(2x + 3)

x = 1, 8

x = – , –1

2

9

2

x = – , 7 5

2

x = – , 5 3

2

26

Topic7.1.1


Solution follows…


Solve the following equations.

15. y(2y + 7) = 9(2y + 7)

16. (x + 2)(x – 2) = 3x

17. 2x(2x – 5) = 3(2x – 5)

18. 2x(3x – 1) + 7 = 7(2 – 3x)

x = 4, –1

y = – , 9 7

2

x = , 5

2

3

2

x = – , 7

2

1

3

27

Topic7.1.1


Solution follows…


19. The product of two consecutive positive numbers is 30. Find the numbers.

20. The product of two consecutive positive odd numbers is 35. Find the numbers.

21. The area of a rectangle is 35 ft2. If the width of the rectangle is x ft and the length is (3x + 16) ft, find the value of x.

5, 6

5, 7

x =5

3

28

Topic7.1.1


Solution follows…


22. The area of a rectangle is 75 cm2. If the length of the rectangle is (4x + 25) cm and the width is 2x cm, find the dimensions of the rectangle.

23. Eylora has x pet goldfish and Leo has (4x – 25). If the product of the numbers of Eylora’s and Leo’s goldfish is 21, how many goldfish does Leo have?

24. Scott fixed x computers and Meimei fixed (5x – 7) computers. If the product of the number each fixed is 6, who fixed more computers?

Meimei fixed more computers

3 goldfish

cm by 30 cm5

2

29

Topic7.1.1

Round UpRound Up


Don’t forget that you need to rearrange the equation until you’ve got a zero on one side before you can factor a quadratic.

Not all quadratics can be factored like this, as you’ll see in the next Topic.

1 Topic 7.1.1 Solving Quadratic Equations by Factoring Solving Quadratic Equations by Factoring.

Documents