1. 9.1 Square Roots and the Pythagorean Theorem
By definition Ö25 is the number you would multiply times itself to get 25 for an answer.
Because we are familiar with multiplication, we know that Ö25 = 5
Numbers like 25, which have whole numbers for their square roots, are called perfect squares
You need to memorize at least the first 15 perfect squares

2. Perfect square
Square root
Perfect square
Square root 1
Ö1 = 1 81
Ö81 = 9 4
Ö4 = 2
100
Ö100 = 10 9
Ö9 = 3
121
Ö121 = 11 16
Ö16 = 4
144
Ö144 = 12 25
Ö25 = 5
169
Ö169 = 13 36
Ö36 = 6
196
Ö196 = 14 49
Ö49 = 7
225
Ö225 = 15 64
Ö64 = 8

3. 9.1 Square Roots and the Pythagorean Theorem
Every whole number has a square root
Most numbers are not perfect squares, and so their square roots are not whole numbers.
Most numbers that are not perfect squares have square roots that are irrational numbers
Irrational numbers can be represented by decimals that do not terminate and do not repeat
The decimal approximations of whole numbers can be determined using a calculator

4. 9.1 Square Roots and the Pythagorean Theorem
Find the square roots of the given numbers
If the number is not a perfect square, use a calculator to find the answer correct to the nearest hundredth. 81 37
158

5. 9.1 Square Roots and the Pythagorean Theorem
Find the square roots of the given numbers
If the number is not a perfect square, use a calculator to find the answer correct to the nearest thousandth.

6. 9.1 Square Roots and the Pythagorean Theorem
For any right triangle, the sum of the squares of the lengths of the legs a and b, equals the square of the length of the hypotenuse.
a2 + b2 = c2 a b c

7. 9.1 Square Roots and the Pythagorean Theorem
Find c.
a2 + b2 = c2 6 8 c

8. 9.1 Square Roots and the Pythagorean Theorem
Find c.
a2 + b2 = c2 4 6 c

9. 9.1 Square Roots and the Pythagorean Theorem
Find a.
a2 + b2 = c2 a 8 17

10. 9.1 Square Roots and the Pythagorean Theorem
The length of each side of a softball field is 60 feet. How far is it from home to second?
60 ft
= c2
= 7200
c2 = 7200

11. 9.2 Solving Quadratic Equations
Solving x2 = d by Finding Square Roots
If d is positive, then x2 = d has two solutions
The equation x2 = 0 has one solutions:
If d is negative, then x2 = d has no solution.

12. 9.2 Solving Quadratic Equations
Solve
x2 = 49
x2 = 12
x2 = 0
x2 = -9

13. 9.2 Solving Quadratic Equations
Solve
3x2 +1 = 76
4x2 + 6 = 70
5x2 – 7 = 18
3x2 – 10 = 38

14. 9.3 Graphing Quadratic Equations
y = ax2 + bx + c

15. 9.3 Graphing Quadratic Equationsy x
The graph of a quadratic function is a parabola.
Vertex
A parabola can open up or down.
If the parabola opens up, the lowest point is called the vertex.
If the parabola opens down, the vertex is the highest point.
Vertex
NOTE: if the parabola opened left or right it would not be a function!

16. 9.3 Graphing Quadratic Equationsy x
The parabola will open up when the a value is positive.
a < 0
a > 0
The standard form of a quadratic function is
y = ax2 + bx + c
The parabola will open down when the a value is negative.

17. 9.3 Graphing Quadratic Equationsy x
Line of Symmetry
Parabolas have a symmetric property to them.
If we drew a line down the middle of the parabola, we could fold the parabola in half.
We call this line the line of symmetry.
Or, if we graphed one side of the parabola, we could “fold” (or REFLECT) it over, the line of symmetry to graph the other side.
The line of symmetry ALWAYS passes through the vertex.

18. 9.3 Graphing Quadratic Equations
When a quadratic function is in standard form
For example…
Find the line of symmetry of y = 3x2 – 18x + 7
y = ax2 + bx + c,
The equation of the line of symmetry is
Using the formula…
This is best read as …
the opposite of b divided by the quantity of 2 times a.
Thus, the line of symmetry is x = 3.

19. 9.3 Graphing Quadratic Equations
y = –2x2 + 8x –3
We know the line of symmetry always goes through the vertex.
STEP 1: Find the line of symmetry
Thus, the line of symmetry gives us the x – coordinate of the vertex.
STEP 2: Plug the x – value into the original equation to find the y value.
y = –2(4)+ 8(2) –3
y = –2(2)2 + 8(2) –3
To find the y – coordinate of the vertex, we need to plug the x – value into the original equation.
y = –8+ 16 –3
y = 5
Therefore, the vertex is (2 , 5)

20. 9.3 Graphing Quadratic Equations
The standard form of a quadratic function is given by y = ax2 + bx + c
STEP 1: Find the line of symmetry
STEP 2: Find the vertex
STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve.

21. Thus the line of symmetry is x = 1
9.3 Graphing Quadratic Equations y x
Let's Graph ONE! Try …
y = 2x2 – 4x – 1
STEP 1: Find the line of symmetry
Thus the line of symmetry is x = 1

22. 9.3 Graphing Quadratic Equationsy x
Let's Graph ONE! Try …
y = 2x2 – 4x – 1
STEP 2: Find the vertex
Since the x – value of the vertex is given by the line of symmetry, we need to plug in x = 1 to find the y – value of the vertex.
Thus the vertex is (1 ,–3).

23. 9.3 Graphing Quadratic Equationsy x
Let's Graph ONE! Try …
y = 2x2 – 4x – 1
STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve. x y 2 -1 3 5

24. 9.3 Graphs of Quadratic Equations
For y = -x2 - 2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.
-Vertex:
x =(-b/2a)
x= -(-2/2(-1))
x= 2/(-2)
x= -1
Solve for y:
y = -x2 -2x + 8
y = -(-1)2 -(2)(-1) + 8
y = -(1)
y = 9
Vertex is (-1, 9)

25. 9.4 The Quadratic Formula
The quadratic formula allows you to find the roots of a quadratic equation (if they exist) even if the quadratic equation does not factorise.
The formula states that for a quadratic equation of the form :
ax2 + bx + c = 0
The roots of the quadratic equation are given by :

26. The Quadratic Formula.

27. Example 1
Use the quadratic formula to solve the equation :
x 2 + 5x + 6 = 0
Solution:
x2 + 5x + 6= 0
a = 1 b = 5 c = 6
x = - 2 or x = - 3
These are the roots of the equation.

28. Example 3
Use the quadratic formula to solve the equation :
8x2 - 22x + 15= 0
Solution:
8x2 - 22x + 15= 0
a = 8 b = c = 15
x = 3/2 or x = 5/4
These are the roots of the equation.

29. Example 4
Use the quadratic formula to solve for x to 2 decimal places.
2x2 + 3x - 7= 0
Solution:
2x 2 + 3x – 7 = 0
a = 2 b = 3 c = - 7
x = or x =
These are the roots of the equation.

30. 9.5 Problem Solving Using the Discriminantac b x 2 4 - =

31. 9.5 Problem Solving Using the Discriminant
The number of solutions in a quadratic equation
Consider the equation ax2 + bx + c = 0
If b2 – 4ac > 0, then the equation has 2 solutions.
If b2 – 4ac = 0, then the equation has 1 solution.
If b2 – 4ac < 0, then the equation has no solution.

32. 9.5 Problem Solving Using the Discriminant
Find the discriminant of 3x2 + x – 2 = 0 and tell the nature of its roots.
Discriminant = b2 – 4ac = 12 – 4(3)(-2) = 1 – (-24) = = 25
So, there are two solutions

33. 9.5 Problem Solving Using the Discriminant
Determine the number of solutions
2x2 – x + 3 = 0
x2 + x + 4 = 0
3x2 –5x - 3 = 0
2x2 – x - 9 = 0

34. 9.5 Problem Solving Using the Discriminant
Match the discriminant with the graph
b2 – 4ac = 7 b. b2 – 4ac = -2 c. b2 – 4ac = 0

35. 9.6 Graphing Quadratic Inequalities
Steps for Drawing the Graph of an Inequality in Two Variables
Draw the graph of the equation obtained by replacing the inequality sign by an equal sign. Use a dashed line if the inequality is < or >. Use a solid line if the inequality is ≤ or ≥.
Check a point in each of the two regions of the plane determined by the graph of the equation. If the point satisfies the inequality, then shade the region containing the point.

36. 9.6 Graphing Quadratic Inequalities
Solve y > x2