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Square Roots and
Solving Quadratics
with Square Roots
Review 9.1-9.2
 GET YOUR COMMUNICATORS!!!!
Warm Up
Simplify.
1. 52
25
2. 82
3. 122
144
4. 152
5. 202
400
64
225
Perfect Square
 A number that is the square of a
whole number
 Can be represented by
arranging objects in a square.
Perfect Squares
Perfect Squares
1x1=1
2x2=4
3x3=9
 4 x 4 = 16
Perfect Squares
1x1=1
2x2=4
3x3=9
 4 x 4 = 16
Activity:
Calculate the perfect
squares up to 152…
Perfect Squares
1x1=1
 9 x 9 = 81
2x2=4
 10 x 10 = 100
3x3=9
 11 x 11 = 121
 4 x 4 = 16
 12 x 12 = 144
 5 x 5 = 25
 13 x 13 = 169
 6 x 6 = 36
 14 x 14 = 196
 7 x 7 = 49
 15 x 15 = 225
 8 x 8 = 64
Activity:
Identify the following numbers
as perfect squares or not.
i.
ii.
iii.
iv.
v.
vi.
16
15
146
300
324
729
Activity:
Identify the following numbers
as perfect squares or not.
16 = 4 x 4
ii. 15
iii. 146
iv. 300
v. 324 = 18 x 18
vi. 729 = 27 x 27
i.
Perfect Squares: Numbers whose square roots
are integers or quotients of integers.
1 1
4 2
9 3
16  4
25  5
36  6
49  7
64  8
81  9
100  10
121  11
144  12
169  13
Perfect Squares
 One property of a perfect
4cm
4cm
16 cm2
square is that it can be
represented by a square
array.
 Each small square in the array
shown has a side length of
1cm.
 The large square has a side
length of 4 cm.
Perfect Squares
 The large square has an area
of 4cm x 4cm = 16 cm2.
4cm
4cm
16 cm2
 The number 4 is called the
square root of 16.
 We write: 4 =
16
Square Root
 A number which, when
multiplied by itself, results in
another number.
 Ex: 5 is the square root of 25.
5 =
25
Finding Square Roots
 We can think “what” times “what”
equals the larger number.
36
=
-6
6
___
x
-6
6
___
Is there another answer?
SO ±6 IS THE SQUARE ROOT OF 36
Finding Square Roots
 We can think “what” times “what”
equals the larger number.
256
16
-16
16
= ___
x -16
___
Is there another answer?
SO ±16 IS THE SQUARE ROOT OF 256
Estimating
Square Roots
25 = ?
Estimating
Square Roots
25 = ±5
Estimating
Square Roots
- 49 = ?
Estimating
Square Roots
- 49 = -7
IF THERE IS A SIGN OUT FRONT OF THE RADICAL
THAT IS THE SIGN WE USE!!
Estimating
Square Roots
27 = ?
Estimating
Square Roots
27 = ?27
Since 27 is not a perfect square, we
will leave it in a radical because that
is an EXACT ANSWER.
If you put
in your calculator it would
give you 5.196, which is a decimal
apporximation.
Estimating
Square Roots
Not all numbers are perfect
squares.
Not every number has an Integer
for a square root.
We have to estimate square roots
for numbers between perfect
squares.
Estimating
Square Roots
 To calculate the square root of a
non-perfect square
1. Place the values of the adjacent
perfect squares on a number line.
2. Interpolate between the points to
estimate to the nearest tenth.
Estimating
Square Roots
 Example:
What are the perfect squares on
each side of 27?
25
30
35 36
27
Estimating
Square Roots
 Example:
half
5
25
30
27
6
35 36
27
Estimate
27 = 5.2
Estimating
Square Roots
 Example:
 Estimate:
27
= 5.2
 Check: (5.2) (5.2) = 27.04
27
Find the two square roots of each number.
A. 49
7 is a square root, since 7 • 7 = 49.
49 = 7
49 = –7 –7 is also a square root, since –7 • –7 = 49.
B. 100
100 = 10
10 is a square root, since 10 • 10 = 100.
100 = –10
–10 is also a square root, since –10 • –10 = 100.
C. 225
225 = 15
15 is a square root, since 15 • 15 = 225.
225 = –15
–15 is also a square root, since –15 • –15 = 225.
Find the two square roots of each number.
A. 25
25 = 5
25 = –5
B. 144
144 = 12
5 is a square root, since 5 • 5 = 25.
–5 is also a square root, since
–5 • –5 = 25.
12 is a square root, since 12 • 12 = 144.
144 = –12 –12 is also a square root, since
–12 • –12 = 144.
C. 289
289 = 17
17 is a square root, since 17 • 17 = 289.
289 = –17 –17 is also a square root, since
–17 • –17 = 289.
Evaluate a Radical
Expression
EXAMPLE SHOWN BELOW
Evaluate b 2  4ac when a  1, b  2, and c  3.
b 2  4ac  (2) 2  4(1)( 3)  4  4(3)
 4  12  16  4
Evaluate
a
Radical
#1
Expression
Evaluate b 2  4ac when a  3, b  6, and c  3.
b  4ac  (6)  4(3)(3)  36  4(9)
2

2
 36  36  0  0
Evaluate
a
Radical
#2
Expression
Evaluate b 2  4ac when a  5, b  8, and c  3.
b  4ac  (8)  4(5)(3)  64  4(15)
2

2
 64  60  4  2
Evaluate
a
Radical
#3
Expression
Evaluate b 2  4ac when a  4, b  9, and c  5.

b  4ac  (9)  4(4)(5)  81  4(20)
2
2
 81  80  1  1
Evaluate
a
Radical
#4
Expression
Evaluate b 2  4ac when a  2, b  9, and c  5.

b 2  4ac  (9) 2  4(2)(5)  81  4(10)
 81  (40)  121  11
SOLVING EQUATIONS
 SOLVING MEANS “ISOLATE” THE
VARIABLE
 x = ???
y = ???
Solving quadratics
 Solve each equation.
SQUARE ROOT BOTH SIDES
a. x2 = 4
b. x2 = 5
c. x2 = 0
x  4
x2  5
x2  0
2
x  2

x 5

x 0

d. x2 = -1
x 2  1
NO SOLUTION
Solve
 Solve 3x2 – 48 = 0
+48 +48
3x2 = 48
3
3
x2 = 16
x 2  16
x  4

Example 1:
 Solve the equation:
1.) x2 – 7 = 9
+7
x2
2.) z2 + 13 = 5
- 13 - 13
+7
z2 = -8
= 16
z 2  8
NO SOLUTION
x 2  16
x  4

Example 2:
 Solve 9m2 = 169
9
m2
9
=
169
9
m 2  169 9

x  169 9
Example 3:
 Solve 2x2 + 5 = 15
-5
-5
2x2 = 10
2
2
x2 = 5
x2  5
x 5
Example:
2. 5 x 2  125
1. 3x  108
2
3
5
3
x2 = 36
x2 = 25
x 2  36
x  6

5
x 2  25
x  5

Example:
3.
4 x  6  42
2
+6
4x2 = 48
4
4
x2 = 12
x 2  12
x  12
+6
Examples:
2
x
5.
 5  21
4
+5 +5
4. 3  5 x  9
-3
-3
2
-5x2 = -12
-5
x2
4
 26 4
4
= 12/5
x2 = 104
x  12 5
x 2  104
-5
x2
2
x  12 5

x  104