1. Welcome! Grab a set of interactive notes and study Guide
Section 5: Quadratic Functions Part 1 Topics 5 - 6
Topics 5
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Grab a set of interactive notes
and study Guide
Homework Assignment:
Algebra Nation Topics 9 & 10
Khan Academy
Topics 6

2. Section 5: Quadratic Functions Part 1 Topics 5 & 6
You will:
write equivalent expressions by factoring
solve a simple quadratic equation by inspection or by taking square roots.

3. Let’s Recall Section 5: Quadratic Functions Part 1 Topics 5 - 6
Find each product.
1. (x + 2)(x + 7) 2. (x – 11)(x + 5)
3. (x – 10)2
Factor each polynomial.
4. x2 + 12x x2 + 2x – 63
6. x2 – 10x x2 – 16x + 32
x2 + 9x + 14
x2 – 6x – 55
x2 – 20x + 100
(x + 5)(x + 7)
(x – 7)(x + 9)
(x – 2)(x – 8)
2(x – 4)2

4. Section 5: Topic 5 Solving Other Quadratics by Factoring
Binomial
A binomial is a polynomial with two terms
Product
Product means the result we get after multiplying.
In Algebra xy means x multiplied by y
(a+b)(a−b) means (a+b) multiplied by (a−b).

5. Section 5: Topic 5 Solving Other Quadratics by Factoring
Special Binomial Products
So when we multiply binomials we get ... Binomial Products!
And we will look at three special cases of multiplying binomials ... so they are Special Binomial Products.

6. Section 5: Topic 5 Solving Other Quadratics by Factoring
Multiplying a Binomial by Itself
What happens when we square a binomial (in other words, multiply it by itself)
(a+b)2 = (a+b)(a+b) =

7. Section 5: Topic 5 Solving Other Quadratics by Factoring
The result:
(a+b)2 = a2 + 2ab + b2
Example: (y+1)2

8. (y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1
Section 5: Topic 5 Solving Other Quadratics by Factoring
Example: (y+1)2
We can use the (a+b)2 case where "a" is y, and "b" is 1:
(y+1)2 = (y)2 + 2(y)(1) + (1)2 = y2 + 2y + 1

9. Section 5: Topic 5 Solving Other Quadratics by Factoring
Example: (y+1)2

10. Section 5: Topic 5 Solving Other Quadratics by Factoring
2. Subtract Times Subtract
And what happens when we square a binomial with a minus inside?
(a−b)2 = (a−b)(a−b)

11. Section 5: Topic 5 Solving Other Quadratics by Factoring
The result:
(a−b)2 = a2 − 2ab + b2

12. Section 5: Topic 5 Solving Other Quadratics by Factoring
Example: (3x−4)2
We can use the (a-b)2 case where "a" is 3x, and "b" is 4:

13. Section 5: Topic 5 Solving Other Quadratics by Factoring
Example: (3x−4)2
We can use the (a-b)2 case where "a" is 3x, and "b" is 4:
(3x−4)2 =
(3x)2 − 2(3x)(4) + (4)2 =
9x2 − 24x + 16

14. Section 5: Topic 5 Solving Other Quadratics by Factoring
3. Difference of Squares
Two terms that are squared and separated by a subtraction sign like this: a2 - b Useful because it can be factored into (a+b)(a−b)
x2 – 9 =
  4x2 − 9 =

15. Section 5: Topic 5 Solving Other Quadratics by Factoring
Difference of Squares
Two terms that are squared and separated by a subtraction sign like this: a2 - b2 x2 – 9 = (x-3)(x+3)
  4x2 − 9 = (2x)2 − (3)2 = (2x+3)(2x−3)

16. Section 5: Topic 5 Solving Other Quadratics by Factoring
The Three Cases:
Remember these patterns, they will save you time.

17. Section 5: Topic 5 Solving Other Quadratics by Factoring
Let’s Recall
Determine whether the following are perfect squares. If so, find the square root. 64
2. 36
yes; 6
yes; 8
3. 45 no
4. x2
yes; x
5. y8
yes; y4
6. 4x2
yes; 2x
7. 9y7 no
8. 49p10
yes;7p5

18. Section 5: Topic 5 Perfect Square Trinomials

19. Section 5: Topic 5 Perfect Square Trinomials
A trinomial is a perfect square if:
• The first and last terms are perfect squares.
• The middle term is two times one factor from the first term and one factor from the last term.
9x x
(3x) 2
2(3x (2)
(2) 2 )

20. Example 1A: Recognizing and Factoring Perfect-Square Trinomials
Section 5: Topic 5 Perfect Square Trinomials
Example 1A: Recognizing and Factoring Perfect-Square Trinomials
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
9x2 – 15x + 64
9x2 – 15x + 64
(8) 2
(3x) 2
2(3x 8) 
2(3x 8) ≠ –15x. 
9x2 – 15x + 64 is not a perfect-square trinomial because –15x ≠ 2(3x  8).

21. Example 1B: Recognizing and Factoring Perfect-Square Trinomials
Section 5: Topic 5 Perfect Square Trinomials
Example 1B: Recognizing and Factoring Perfect-Square Trinomials
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
81x2 + 90x + 25
81x2 + 90x + 25
(5) 2
(9x) 2
2(9x 5) ●
The trinomial is a perfect square. Factor.

22. Section 5: Topic 5 Perfect Square Trinomials
Example 1B Continued
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
Method 2 Use the rule.
81x x + 25
a = 9x, b = 5
(9x)2 + 2(9x)(5) + 52
Write the trinomial as a2 + 2ab + b2.
Write the trinomial as (a + b)2.
(9x + 5)2

23. Section 5: Topic 5 Perfect Square Trinomials
Check It Out! Example 1b
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
x2 – 14x + 49
x2 – 14x + 49
(x) 2
2(x 7)
(7) 2 
The trinomial is a perfect square. Factor.

24. Check It Out! Example 1b Continued
Section 5: Topic 5 Perfect Square Trinomials
Check It Out! Example 1b Continued
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
Method 2 Use the rule.
x2 – 14x + 49
a = 1, b = 7
(x)2 – 2(x)(7) + 72
Write the trinomial as a2 – 2ab + b2.
(x – 7)2
Write the trinomial as (a – b)2.

25. Section 5: Topic 5 Perfect Square Trinomials
Check It Out! Example 1c
Determine whether each trinomial is a perfect square. If so, factor. If not explain.
9x2 – 6x + 4
9x –6x
(3x) 2
2(3x 2)
(2) 2 
2(3x)(4) ≠ – 6x
9x2 – 6x + 4 is not a perfect-square trinomial because –6x ≠ 2(3x 2) 

26. Section 5: Topic 5 Difference of squares

27. Section 5: Topic 5 Difference of squares
The difference of two squares, a2 – b2can be written as the product (a + b)(a – b). You can use this pattern to factor some polynomials.
A polynomial is a difference of two squares if:
There are two terms, one subtracted from the other.
Both terms are perfect squares.
4x2 – 9
(2x) 2 (3) 2 

28. Example 3B: Recognizing and Factoring the Difference of Two Squares
Section 5: Topic 5 Difference of squares
Example 3B: Recognizing and Factoring the Difference of Two Squares
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
100x2 – 4y2
100x2 – 4y2
(2y) 2 
(10x) 2
The polynomial is a difference of two squares.
a = 10x, b = 2y
(10x + 2y)(10x – 2y)
Write the polynomial as (a + b)(a – b).
100x2 – 4y2 = (10x + 2y)(10x – 2y)

29. Section 5: Topic 5 Difference of squares
Check It Out! Example 3a
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
1 – 4x2
1 – 4x2
(2x) 2 
(1) 2
The polynomial is a difference of two squares.
a = 1, b = 2x
(1 + 2x)(1 – 2x)
Write the polynomial as (a + b)(a – b).
1 – 4x2 = (1 + 2x)(1 – 2x)

30. Section 5: Topic 5 Difference of squares
Check It Out! Example 3c
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
16x2 – 4y5
16x2 – 4y5
(4x) 2
4y5 is not a perfect square.
16x2 – 4y5 is not the difference of two squares because 4y5 is not a perfect square.

31. Let’s Recall Find each square root. 1. 6 2. 11 3.
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Let’s Recall
Find each square root. 1. 6 2. 11 3.

32. Objective Solve quadratic equations by using square roots.
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Objective
Solve quadratic equations by using square roots.

33. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Some quadratic equations cannot be easily solved by factoring.
Square roots can be used to solve some of these quadratic equations.
Recall every positive real number has two square roots, one positive and one negative.

34. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Positive
Square root of 9
Negative
Square root of 9
When you take the square root of a positive number and the sign of the square root is not indicated, you must find both the positive and negative square root. This is indicated by ±√
Positive and negative
Square roots of 9

35. Section 5: Topic 6 Solving Quadratics by Taking Square Roots

36.   Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Example 1A: Using Square Roots to Solve x2 = a
Solve using square roots. Check your answer.
x2 = 169
Solve for x by taking the square root of both sides. Use ± to show both square roots.
x = ± 13
The solutions are 13 and –13.
Check x2 = 169
(13) 
x2 = 169
(–13) 
Substitute 13 and –13 into the original equation.

37. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Check It Out! Example 1c
Solve using square roots. Check your answer.
x2 = –16
There is no real number whose square is negative.
There is no real solution.

38. Example 2A: Using Square Roots to Solve Quadratic Equations
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Example 2A: Using Square Roots to Solve Quadratic Equations
Solve using square roots.
x2 + 7 = 7
–7 –7
x2 + 7 = 7
x2 = 0
Subtract 7 from both sides.
Take the square root of both sides.
The solution is 0.

39. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Check It Out! Example 2b
Solve by using square roots. Check your answer.
(x – 5)2 = 16
(x – 5)2 = 16
Take the square root of both sides. Use ± to show both square roots.
x – 5 = ±4
x – 5 = 4 or x – 5 = –4
Write two equations, using both the positive and negative square roots, and solve each equation.
x = 9 or x = 1
The solutions are 9 and 1.

40. Check It Out! Example 2b Continued
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Check It Out! Example 2b Continued
Solve by using square roots. Check your answer.
(x – 5)2 = 16
(1 – 5)
(– 4)
Check (x – 5)2 = 16
(9 – 5)  

41. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
When solving quadratic equations by using square roots, you may need to find the square root of a number that is not a perfect square.
In this case, the answer is an irrational number. You can approximate the solutions.

42. Example 3A: Approximating Solutions
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Example 3A: Approximating Solutions
Solve. Round to the nearest hundredth.
x2 = 15
Take the square root of both sides.
x  3.87
Evaluate on a calculator.
The approximate solutions are 3.87 and –3.87.

43. Example 3B: Approximating Solutions
Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Example 3B: Approximating Solutions
Solve. Round to the nearest hundredth.
–3x = 0
–3x = 0
–90 –90
Subtract 90 from both sides.
Divide by – 3 on both sides.
x2 = 30
Take the square root of both sides.
x  5.48
Evaluate on a calculator.
The approximate solutions are 5.48 and –5.48.

44. Section 5: Topic 6 Solving Quadratics by Taking Square Roots
Check It Out! Example 3b
Solve. Round to the nearest hundredth.
2x2 – 64 = 0
2x2 – 64 = 0
Add 64 to both sides.
Divide by 2 on both sides.
x2 = 32
Take the square root of both sides.
The approximate solutions are 5.66 and –5.66.

45. Section 5: Quadratic Functions Part 1 Topics 5 & 6
Homework
Algebra Nation 9 & 10
Khan Academy