4 1 radicals and pythagorean theorem

Radicals and Pythagorean Theorem

From here on we need a scientific calculator.
You may use the digital calculators on your personal devices.

Make sure the calculator could displace your input so that
you may check the input before executing it.

Square Root

“9 is the square of 3” may be rephrased backwards as
“3 is the square root of 9”.
Square Root

Definition: If a is > 0, and a2 = x, then we say a is the square
root of x. This is written as sqrt(x) = a, or x = a.
Square Root

Example A.
a. Sqrt(16) =
c.–3 =
b. 1/9 =
d. 3 =
Square Root

Example A.
a. Sqrt(16) = 4
c.–3 =
b. 1/9 =
d. 3 =
Square Root

Example A.
a. Sqrt(16) = 4
c.–3 =
b. 1/9 = 1/3
d. 3 =
Square Root

Example A.
a. Sqrt(16) = 4
c.–3 = doesn’t exist
b. 1/9 = 1/3
d. 3 =
Square Root

Example A.
a. Sqrt(16) = 4
b. 1/9 = 1/3
d. 3 = 1.732.. (calculator)
Square Root

Example A.
a. Sqrt(16) = 4
b. 1/9 = 1/3
d. 3 = 1.732.. (calculator)
Note that the square of both +3 and –3 is 9,
but we designate sqrt(9) or 9 to be +3.
Square Root

Example A.
a. Sqrt(16) = 4
b. 1/9 = 1/3
d. 3 = 1.732.. (calculator)
Note that the square of both +3 and –3 is 9,
but we designate sqrt(9) or 9 to be +3.
We say “–3” is the “negative of the square root of 9”.
Square Root

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
Following are the square numbers and square-roots that one
needs to memorize.

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
needs to memorize. These numbers are special because
many mathematics exercises utilize square numbers.

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
We may estimate the sqrt
of other small numbers using
this table.

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
this table. For example,
25 < 30 < 36

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
25 < 30 < 36
hence
25 < 30 <36

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
25 < 30 < 36
hence
25 < 30 <36
or 5 < 30 < 6

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
25 < 30 < 36
hence
25 < 30 <36
or 5 < 30 < 6
Since 30 is about half way
between 25 and 36,

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
25 < 30 < 36
hence
25 < 30 <36
or 5 < 30 < 6
between 25 and 36,
so we estimate that30  5.5.

0 02 = 0 0 = 0
1 12 = 1 1 = 1
2 22 = 4 4 = 2
3 32 = 9 9 = 3
4 42 = 16 16 = 4
5 52 = 25 25 = 5
6 62 = 36 36 = 6
7 72 = 49 49 = 7
8 82 = 64 64 = 8
9 92 = 81 81 = 9
10 102 = 100 100 = 10
11 112 = 121 121 = 11
25 < 30 < 36
hence
25 < 30 <36
or 5 < 30 < 6
between 25 and 36,
so we estimate that30  5.5.
In fact 30  5.47722….

Equations of the form x2 = c has two answers:
x = +c or –c if c>0.

x = +c or –c if c>0.
Example B. Solve the following equations.
a. x2 = 25

x = +c or –c if c>0.
a. x2 = 25
x = ±25

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
Solution does not exist.

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
c. x2 = 8

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
c. x2 = 8
x = ±8

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
c. x2 = 8
x = ±8  ±2.8284.. by calculator

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
c. x2 = 8
x = ±8  ±2.8284.. by calculator
exact answer approximate answer

x = +c or –c if c>0.
a. x2 = 25
x = ±25 = ±5
b. x2 = –4
c. x2 = 8
x = ±8  ±2.8284.. by calculator
exact answer approximate answer
In geometry square roots show up in distance problems
because of Pythagorean theorem.

A right triangle is a triangle with a right angle as one of its
angle.

angles. The longest side C of a right triangle is called the
hypotenuse,
hypotenuse
C

hypotenuse, the two sides A and B forming the right angle
are called the legs.
hypotenuse
legs
A
B
C

hypotenuse, the two sides A and B forming the right angle
are called the legs.
Pythagorean Theorem
Given a right triangle with labeling as shown, then
A2 + B2 = C2
hypotenuse
legs
A
B
C

There are two types of problems that use the Pythagorean
Theorem to find the sides of the right triangles

Theorem to find the sides of the right triangles-finding the
hypotenuse versus finding the legs.

Example C. Find the missing side of the following right
triangles. Find the exact answer and the approximate answer.
Draw.
a. a = 5, b = 12, c = ?

Draw.
a. a = 5, b = 12, c = ?
Since it is a right triangle,
122 + 52 = c2

Draw.
a. a = 5, b = 12, c = ?
122 + 52 = c2
144 + 25 = c2

Draw.
a. a = 5, b = 12, c = ?
122 + 52 = c2
144 + 25 = c2
169 = c2

Draw.
a. a = 5, b = 12, c = ?
122 + 52 = c2
144 + 25 = c2
169 = c2
So c = ±169 = ±13

Draw.
a. a = 5, b = 12, c = ?
122 + 52 = c2
144 + 25 = c2
169 = c2
So c = ±169 = ±13
Since length can’t be
negative, therefore c = 13.

b. a = 5, c = 12, b = ?

b. a = 5, c = 12, b = ?
b2 + 52 = 122

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
But length can’t be negative,
therefore b = 119  10.9

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
The Distance Formula

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
Let (x1, y1) and (x2, y2) be two
points, D = distance between them,
(x1, y1)
(x2, y2)
D

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
then D2 = Δx2 + Δy2
(x1, y1)
(x2, y2)
Δy
Δx
D

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
then D2 = Δx2 + Δy2 where
(x1, y1)
(x2, y2)
Δx = x2 – x1
Δy
Δx = x2 – x1
D

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
(x1, y1)
(x2, y2)
Δx = x2 – x1 Δy = y2 – y1and
Δy = y2 – y1
Δx = x2 – x1
D
by the Pythagorean Theorem.

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
(x1, y1)
(x2, y2)
Δx = x2 – x1 Δy = y2 – y1and
Δy = y2 – y1
Δx = x2 – x1by the Pythagorean Theorem.
Hence we’ve the Distant Formula:
D = √ Δx2 + Δy2
D

b. a = 5, c = 12, b = ?
b2 + 52 = 122
b2 + 25 = 144
b2 = 144 – 25
So b = ±119  ±10.9.
(x1, y1)
(x2, y2)
Δx = x2 – x1 Δy = y2 – y1and
Δy = y2 – y1
Δx = x2 – x1by the Pythagorean Theorem.
Hence we’ve the Distant Formula:
D = √ Δx2 + Δy2 = √ (x2 – x1)2 + (y2 – y1)2
D

Example D. Find the distance between (–1, 3) and (2, –4).
(-1, 3)
– ( 2, -4)
D
–3, 7
Δx Δy

(-1, 3)
– ( 2, -4)
D
7
–3
–3, 7
Δx Δy

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root
If r3 = x, then we say a is the cube root of x.

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root
If r3 = x, then we say a is the cube root of x. We write this as
x = r.3

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root
x = r. In general, if r k = x, then we say r is the k’th root of x,3

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root
x = r. In general, if r k = x, then we say r is the k’th root of x,
and we write it as a = x.
3
k

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
Higher Root
and we write it as a = x. In the cases of even roots,
i.e. k = 2, 4, 6, … we must have x > 0 and that x = a > 0.
3
k
k

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 =
3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2
3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 =
3 3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1
3 3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 =
3 3 3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
3 3 3
Higher Root

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
d. 16 =
3 3 3
Higher Root
4

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
d. 16 = 2
3 3 3
Higher Root
4

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
d. 16 = 2 e. –16 =
3 3 3
Higher Root
4 4

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
d. 16 = 2 e. –16 = not real
3 3 3
Higher Root
4 4

(-1, 3)
– ( 2, -4)
D = (–3)2 + 72 = 58  7.62
D
7
–3
–3, 7
Δx Δy
3
k
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
d. 16 = 2 e. –16 = not real f. 10 ≈ 2.15..
3 3 3
Higher Root
4 4 3

Rational and Irrational Numbers

The number 2 is the length of the
hypotenuse of the right triangle as shown.
2
1
1

2
1
1
It can be shown that 2 can not be
represented as a ratio of whole numbers i.e.
P/Q, where P and Q are integers.

2
1
1
Hence these numbers are called irrational (non–ratio)
numbers.

2
1
1
numbers. Most real numbers are irrational, not fractions, i.e.
they can’t be represented as ratios of two integers.

2
1
1
they can’t be represented as ratios of two integers. The real
line is populated sparsely by fractional locations.

2
1
1
line is populated sparsely by fractional locations. The
Pythagorean school of the ancient Greeks had believed that
all the measurable quantities in the universe are fractional
quantities. The “discovery” of these extra irrational numbers
caused a profound intellectual crisis.

2
1
1
line is populated sparsely by fractional locations. The
Pythagorean school of the ancient Greeks had believed that
all the measurable quantities in the universe are fractional
quantities. The “discovery” of these extra irrational numbers
caused a profound intellectual crisis. It wasn’t until the last two
centuries that mathematicians clarified the strange questions
“How many and what kind of numbers are there?”

Exercise A. Solve for x. Give both the exact and approximate
answers. If the answer does not exist, state so.
1. x2 = 1 2. x2 – 5 = 4 3. x2 + 5 = 4
4. 2x2 = 31 5. 4x2 – 5 = 4 6. 5 = 3x2 + 1
7. 4x2 = 1 8. x2 – 32 = 42 9. x2 + 62 = 102
10. 2x2 + 7 = 11 11. 2x2 – 5 = 6 12. 4 = 3x2 + 5
x
3
4
Exercise B. Solve for x. Give both the exact and approximate
answers. If the answer does not exist, state so.
13. 4
3
x14. x
12
515.
x
1
116. 2
1
x17. 3 2
3
x18.

x
4
19.
x
x20.
3 /3
21.
43 5 2
6 /3
Exercise C. Given the following information, find the rise and
run from A to B i.e. Δx and Δy. Find the distance from A to B.
A
22.
B
A
23.
B
24. A = (2, –3) , B = (1, 5) 25. A = (1, 5) , B = (2, –3)
26. A = (–2 , –5) , B = (3, –2) 27. A = (–4 , –1) , B = (2, –3)
28. Why is the distance from A to B the same as from B to A?

Exercise D. Find the exact answer.
3
–129. 30. 13
–12534.
8 31. –13
8
3
–2732.
33. –13
64
3
10035. 100036.
3
10,00037. 1,000,00038.
3
0.0139. 0.00140.
3
0.000141. 0.00000142.
3

4 1 radicals and pythagorean theorem

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4 1 radicals and pythagorean theorem