1. Quadratic and Higher Degree Equations and Functions
Algebra II Chapter 9
Quadratic and Higher Degree Equations and Functions
9.1
9.2
9.3
9.4
9.5
9.6

2. Solving Quadratic Equations by Completing the Square
9.1
Solving Quadratic Equations by Completing the Square
Objectives:
Solve quadratic equations by:
Square root property
Completing the square

3. We previously have used factoring to solve quadratic equations.
This chapter will introduce additional methods for solving quadratic equations.
Square Root Property
If b is a real number and if a2 = b, then

4. Example Use the square root property to solve. x2 = 49 2x2 − 4 = 0
y = 1 or 5

5. Example Use the square root property to solve x2 + 4 = 0 x2 = 4

6. Completing the Square x2 – 10x x2 + 16x x2 – 7x = x2 – 10x + 25
What constant term should be added to the following expressions to create a perfect square trinomial?
x2 – 10x
add 52 = 25
x2 + 16x
add 82 = 64
x2 – 7x
= x2 – 10x + 25
= x2 + 16x + 64
= x2 – 7x + 49/4
= (x – 5)2
= (x + 8)2
= (x – 7/2)2
add

7. Solving a Quadratic Equation in x by Completing the Square
The coefficient of x2 needs to be 1. Divide both sides of the equation by the coefficient of x2.
Isolate all variable terms on one side of the equation.
Complete the square for the resulting binomial by adding the (half of the coefficient of x)2 to both sides of the equation.
Factor the resulting perfect square trinomial and write it as the square of a binomial.
Use the square root property to solve for x.

8. Example Solve y2 + 6y = 8 by completing the square.
y = 4 or 2

9. Example (y + ½)2 = Solve y2 + y – 7 = 0 by completing the square.

10. Example Solve 2x2 + 14x – 1 = 0 by completing the square.

11. Summary 9.1 Objectives: Solve quadratic equations by:
Square root property
Completing the square

12. Solving Quadratic Equations by the Quadratic Formula
9.2
Solving Quadratic Equations by the Quadratic Formula
Objectives:
Solve quadratic equations by Quadratic Formula
Determine the number and type of solutions of a quadratic equation by using the discriminant

13. Another technique for solving quadratic equations is to use the quadratic formula.
The formula is derived from completing the square of a general quadratic equation.
TEN BONUS points in the test category for proving the quadratic formula.
See the next slide for how to do this

14. Quadratic Formula Proof

15. Quadratic Formula
A quadratic equation written in the form, ax2 + bx + c = 0, has the solutions

16. Example Solve 11n2 – 9n = 1 for n. 11n2 – 9n – 1 = 0
a = 11, b = −9, c = −1

17. Example
Solve x2 + x – = 0.
x2 + 8x – 20 = 0
a = 1, b = 8, c = −20

18. Example Solve x(x + 6) = 30 by the quadratic formula.
a = 1, b = 6, c = 30

19. Example
Harold throws a ball in the air with the starting velocity of 30 feet per second. The ball is 5 feet high when it leaves his hand. After how many seconds will it hit the ground?
0 = −16t2 + 30t + 5
a = −16, b = 30, and c = 5.
Reject negative time
2.0 seconds to hit the ground.

20. The Discriminant
The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant.
The discriminant will take on a value that is positive, 0, or negative.
The value of the discriminant is
positive – two distinct real solutions
zero – one real solution
negative – two complex solutions (imaginary)

21. Example
Use the discriminant to determine the number and type of solutions for the following equation.
x2 – 7x + 6 = 0
a = 1, b = –7, and c = 6
b2 – 4ac = (−7)2 – 4(1)(6)
= 49 – 24
= 25
There are two real solutions.
5 – 4x + 12x2 = 0
a = 12, b = –4, and c = 5
b2 – 4ac = (–4)2 – 4(12)(5)
= 16 – 240
= –224
There are no real solutions, there are two complex solutions (imaginary).
And talk about the graphs ..
# of x-intercepts, X-int, Y-int, Direction opens

22. Example
Use the discriminant to determine the number and type of solutions for the following equation.
x2 – 6x + 9 = 0
a = 1, b = –6, and c = 9
b2 – 4ac = (−6)2 – 4(1)(9)
= 36 – 36
= 0
There is one real solution.
Talk about the graph ..
# of x-intercepts
X-int
Y-int
Direction opens

23. 9.2 summary Objectives: Solve quadratic equations by Quadratic Formula
Determine the number and type of solutions of a quadratic equation by using the discriminant

24. Solving Equations by Using Quadratic Methods
9.3
Solving Equations by Using Quadratic Methods
Objectives:
Solve various equations that are quadratic in form
Solve problems that lead to quadratic equations

25. Solving a Quadratic Equation
1. If the equation is in the form (ax + b)2 = c, use the square root property and solve.
2. Write the equation in standard form: ax2 + bx + c = 0.
If factorable, then solve the equation by the factoring method.
If a = 1 and b is an even number, then solve by completing the square.
Solve the equation by the quadratic formula.

26. Example Solve: Check: The solution is 8 or the solution set is {8}.
+16 -8 -2
Solve:
-10
We need to isolate the radical and square both sides of the equation.
Check:
True
False
The solution is 8 or the solution set is {8}.
x = 8 or x = 2

27. Example
Solve
LCD: (x – 2)(x+ 2)
Restrictions: x ≠ 2 or –2

28. Example Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1)
Let a = 1, b = –3, c = 1

29. Example
-20 -2
+10
Solve
-2 = 5
= +2 5 +8 5

30. Example Solve 9x4 + 5x2 – 4 = 0. 9u2 + 5u – 4 = 0 (9u – 4)(u + 1) = 0
-36
Solve 9x4 + 5x2 – 4 = 0.
Substitute u = x2 into the equation.
9u2 + 5u – 4 = 0
(9u – 4)(u + 1) = 0
9u – 4 = 0 or u + 1 = 0
u = 4/9 or u = -1
x2 = 4/9 or x2 = -1 +9 -4
+1 = 9 +5 9
By the square root property
The solutions are x = ± or x = ± i.

31. Example x1/3 = Solve 3x2/3 + 11x1/3 = 4. (x1/3)3 = ( )3
Replace x1/3 with u.
3u2 + 11u – 4 = 0
Solve by factoring.
(3u – 1)(u + 4) = 0
3u – 1= u + 4 = 0
u = 1/ u = -4
(x1/3)3 = ( )3
x1/3 = −4
(x1/3)3 = (−4)3
x = −64
Both solutions check, so the solutions are

32. Example
Beach and Fargo are about 400 miles apart. A salesperson travels from Fargo to Beach one day at a certain speed. She returns to Fargo the next day and drives 10 mph faster. Her total time was hours. Find her speed to Beach and the return speed to Fargo. •
Rate
Time
= Distance
To beach
Return to Fargo X
400/X
400
X + 10
400/X+10
400
Reduces by 4
continue

33. 0 = (11x + 60)(x – 50)
The speed to Beach was 50 mph and her return speed to Fargo was 60 mph.
x = -60/11 or x = 50

34. 9.3 summary Solving a Quadratic Equation
Objectives:
Solve various equations that are quadratic in form
Solve problems that lead to quadratic equations
Solving a Quadratic Equation
1. If the equation is in the form (ax + b)2 = c, use the square root property and solve.
Write the equation in standard form:
ax2 + bx + c = 0.
If factorable, then solve the equation by the factoring method.
If a = 1 and b is an even number, then solve by completing the square.
Solve the equation by the quadratic formula.

35. Zeros of Polynomial Functions
9.4
Zeros of Polynomial Functions
Objectives:
Use the Rational Zero Theorem to find rational zeros
Find zeros of a polynomial function

37. Example 1
List all possible rational zeros of f(x) = x3 − 3x2 − 4x + 12.
Possible rational zeros =

38. Example 2 List all possible rational zeros of
f(x) = 8x3 + 10x2 − 11x + 2.
There are 10 possible rational zeros.
The actual rational zeros are

39. Find all the zeros of f(x) = x3 + 2x2 − 5x − 6.
Example 3
Find all the zeros of f(x) = x3 + 2x2 − 5x − 6.
Have the class test which is a zero using synthetic division …
The zero remainder tells us that 2 is a zero of the polynomial function.
Thus x − 2 is a factor of the polynomial. The first three numbers in the bottom row of the synthetic division, 1, 4, 3, give the coefficients of the other factor.
f(x) = x3 + 2x2 – 5x – 6 = (x – 2) (x2 + 4x + 3)
= (x – 2) (x + 1)(x + 3)
The zeros of f are 2, −1, and −3

40. Example 4
Find all the zeros of f(x) = x3 + 7x2 + 11x − 3.
Now we use synthetic division to find the rational zeros
x3 + 7x2 + 11x − 3 = 0
(x + 3)(x2 + 4x −1) = 0
x + 3 = 0 or x2 + 4x −1 = 0
continue

41. (x + 3)(x2 + 4x −1) = 0 x2 + 4x + __ = 1 + __ x2 + 4x + 4 = 1 + 4
I would have solved by completing the square since a = 1 and b is even
(x + 3)(x2 + 4x −1) = 0
x2 + 4x + __ = 1 + __
x2 + 4x + 4 = 1 + 4
(x + 2)2 = 5
x + 2 = +√5
The zeros of f are
Among these three real zeros, one zero is rational, and two are irrational. –

42. Example 5
Find all the zeros of f(x) = –2x5 – 9x4 + 17x3 + 93x2 + 9x − 108 without a graphing calculator, draw a complete picture of f(x). An incomplete graph of f(x) is given to the right.
The graph tells me that ____ and ___ are zeros 1 3 1 3
Other possible rational roots:
±(factors of 36)/(factors of 2)
1,2,3,4,6,9,12,18,36
1,2
Test with syn division to find remainder of 0 -3
f(x) = (x – 1)(x – 3) (-2x3 – 17x2 – 45x – 36 )
= (x – 1)(x – 3) (x + 3) (-2x2 – 11x – 12)
= –(x – 1)(x – 3) (x + 3) (2x2 + 11x + 12)
= –(x – 1)(x – 3) (x + 3)(x + 4) (2x + 3)
The zeros are:
x = 1, x = 3, x = -3, x = -4, and x = -3/2
Then draw a sketch on the whiteboard. Reminder use the end behavior and zeros of the function.

43. 9.4 summary
Objectives:
Use the Rational Zero Theorem to find rational zeros
Possible rational zeros: ±factors of last over factors of first
Find zeros of a polynomial function
Test the rational zeros by using synthetic division

44. The Fundamental Theorem of Algebra
9.5
The Fundamental Theorem of Algebra
Objectives:
Solve polynomial equations
Find polynomials with given zeros

45. If the degree of the polynomial function or equation is 4 or higher, it is often necessary to find more than one linear factor by synthetic division.

46. Solve: x4 – 6x2 – 8x + 24 = 0. (x – 2)(x – 2)(x2 + 4x + 6) = 0
Example 1
Solve: x4 – 6x2 – 8x + 24 = 0. 2
This zero remainder indicates that 2 is a root again.
(x – 2)(x – 2)(x2 + 4x + 6) = 0
continue

47. x4 – 6x2 – 8x + 24 = 0 (x – 2)(x – 2)(x2 + 4x + 6) = 0
x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0
x = or x = or x2 + 4x + 6 = 0
x2 + 4x + __ = –6 + __
The solution of
x4 – 6x2 – 8x + 24 = 0 are
x2 + 4x + 4 = –6 + 4
(x + 2)2 = –2
x + 2= +√–2

48. Properties of Roots of Polynomial Equations
If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots.
If a + bi is a root of a polynomial equation with real coefficients (b ≠ 0), then the imaginary number a – bi is also a root. Imaginary roots, if they exist, occur in conjugate pairs.

49. The Fundamental Theorem of Algebra
If f(x) is a polynomial of degree n, where n 1, then the equation f(x) = 0 has at least one complex root.
The following statement is a result of the Fundamental Theorem of Algebra, and we use it often. If f(x) is a polynomial n, where n 1,then f(x) = 0 has exactly n roots, where roots are counted according to their multiplicity.

50. Example 2
Find a fourth-degree polynomial function f(x) with real coefficients, a leading coefficient of 1, and has 1, 4, and i as zeros.
Zeros are: x = 1, x = 4, x = i,and x = -i
Factors are: (x – 1)(x – 4)(x – i)(x + i)

51. Example 3
Find a third-degree polynomial function f(x) with real coefficients, a leading coefficient of 1, and has ½, -4, and 2 as zeros.
Zeros are: x = ½, x = -4, x = 2
Factors are: (2x – 1)(x + 4)(x – 2)
F(x) = (2x2 + 7x – 4)(x – 2)
F(x) = 2x3 + 3x2 – 18x + 8

52. 9.5 summary Objectives: Solve polynomial equations
Find polynomials with given zeros

53. Nonlinear Inequalities in One Variable
9.6
Nonlinear Inequalities in One Variable
Objectives:
Solve polynomial inequalities of degree 2 or greater
Solve inequalities that contain rational expressions with variables in the denominator

54. Examples We previously have solved linear inequalities.
A quadratic inequality is an inequality that can be written so that one side is a quadratic expression and the other side is 0.
Examples
3x2 – 4 > 0 –2x2 – 5x + 7  0
x2 + 4x – 6  x2 – 3 < 0
A solution of a quadratic inequality is a value of the variable that makes the inequality a true statement.

55. Solve 3x2 + 5x + 2 > 0 x < -1 or x > -2/3 -1 < x < -2/3
We are looking for values of x that will make this a true statement.
If we graph the quadratic equation y = 3x2 + 5x + 2, the points of the parabola that lie above the x-axis would provide values of x where the y-value > 0.
Solve 3x2 + 5x + 2 > 0
To find the x-intercepts:
Find x, when y = 0
3x2 + 5x + 2 = 0
3x2 + 2x + 3x + 2 = 0
x(3x + 2) + 1(3x + 2) = 0
(x + 1)(3x + 2) = 0
x = -1 and -2/3
x < -1 or x > -2/3 -1
-2/3
Solve 3x2 + 5x + 2 < 0
-1 < x < -2/3

56. Example Solve: x(x – 6)(x + 2)  0.
First, solve the equation x(x – 6)(x + 2) = 0.
The solutions are 0, 6 and −2.
End behavior for degree of 3 with positive coefficient is down, up. -2 6
–2 < x < 0 or x  6.

57. Example Solve: x(x – 6)2(x + 2) < 0.
First, solve the equation x(x – 6)2(x + 2) = 0.
The solutions are 0, 6 and −2. (6 has a multiplicity of 2)
End behavior for degree of 4 with positive coefficient is up, up.
Solve: x(x – 6)2(x + 2) < 0.
–2 < x < 0 or x = 6 -2 6
Solve: x(x – 6)2(x + 2) > 0.
x < -2 or 0 < x < 6 or x > 6

58. Solving a Rational Inequality
Solve for values that make all denominators 0.
Solve the related equation.
Separate the number line into regions with the solutions from Steps 1 and 2.
For each region, choose a test point and determine whether its value satisfies the original inequality.
The solution includes the regions whose test point value is a solution. Check whether to include values from Step 2. Be sure not to include values that make any denominator 0.

59. Example Solve Then we need to solve the equality x + 10 = 0 x = −10
Restrictions:
X ≠ 10
Solve
Then we need to solve the equality
-10 10
x + 10 = 0
-11 11
x = −10
-1/-21 > 0
true
10/-10 > 0
false
21/1 > 0
true
x < -10 or x > 10

60. Example Solve p = 3p(p + 4) p = 3p2 + 12p 0 = 3p2 + 11p p(3p + 11) = 0
Restrictions:
P ≠ –4
Solve -4
-11/3 -5
-3.8 -1 1
p = 3p(p + 4)
-1/3 < -3
false
p = 3p2 + 12p
-5/-1 < -15
false
0 = 3p2 + 11p
-3.8/.2 < -11.4
true
1/5 < 3
true
p(3p + 11) = 0
-4 < x < -11/3 or x > 0
p = 0 or p = −11/3

61. 9.6 summary
Objectives:
Solve polynomial inequalities of degree 2 or greater
Solve inequalities that contain rational expressions with variables in the denominator