4 radicals and pythagorean theorem x

Radicals and Pythagorean Theorem

Solutions of equations obtained by the factoring–method
are precise - they are whole numbers or fractions.

Now let’s look at numbers that can’t be easily computed.
We need a scientific calculator to estimate these numbers.

“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,
In fact 30  5.47722….

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,
In fact 30  5.47722….
2 is especially important due
to the Pythagorean Theorem.

Pythagorean Theorem

A right triangle is a triangle with one of its angles a right angle.
Pythagorean Theorem
a right angle
A right triangle

The following is the standard labeling of a right triangle.
Pythagorean Theorem
a right angle
A right triangle

The longest side C, opposite from the right angle of a right
triangle is called the hypotenuse.
hypotenuse
C
Pythagorean Theorem
a right angle
A right triangle

The shorter sides A and B are called the legs.
hypotenuse
legs
A
B
C
Pythagorean Theorem
a right angle
A right triangle

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

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 B. 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

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

x = +c or –c if c>0.
Example C. 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.

Higher Root

3
Higher Root
If r3 = x, then we say a is the cube root of x and we write this
as x = r.

3
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
3 3 3
Higher Root
as x = r.

3
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
3 3 3
Higher Root
as 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
k
Example E.
a. 8 = 2 b. –1 = –1 c.–27 = –3
3 3 3
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 so that x = a > 0.

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
and we write it as a = x. In the cases of even roots,
i.e. k = 2, 4, 6, … we must have x > 0 so that x = a > 0.
4 4 3

Here is one version of the proof of the Pythagorean Theorem.

Given any right triangle, taking four copies of it and arrange
them as shown:

A
B
them as shown:
C
A + B
C

A
B
The area of the large square is the sum of four triangles
and the smaller central squares.
them as shown:
C
A + B
C

A
B
and the smaller central squares.
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B

A
B
and the smaller central squares. Hence in area formulas that:
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2
hence A2 + B2 = C2

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2
hence A2 + B2 = C2
Using this formula for the right triangle with
A = B = 1
1
1

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2
hence A2 + B2 = C2
A = B = 1 so C2 = 12 + 12 = 2,
1
1
C = √2

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2
hence A2 + B2 = C2
A = B = 1 so C2 = 12 + 12 = 2, so that C = √2 ≈ 1.41...
1
1
C = √2

A
B
them as shown:
C
A + B
=
A
B
4 * +
C
C
C
A + B
(A+B)2 = 4* (
AB
2
) + C2, so
A2 + 2AB + B2 = 2AB + C2
hence A2 + B2 = C2
A = B = 1 so C2 = 12 + 12 = 2, so that C = √2 ≈ 1.41...
This number √2 is not a fraction so it's said to be irrational.
1
1
C = √2

The Proof for "√2 is not a fraction"

(The strategy of this proof, to prove an impossibility, is to
exam all the possible forms of p/q if √2 = p/q as a fraction,
then argue that none of these forms is possible.

Hence √2 can’t be a fraction.)

Proof: If √2 = p/q, as a fraction, is even/even, we may reduce it
to the following forms: odd/odd, odd/even or even/odd.

We argue in two parts.
I. If √2 = or
odd
even
odd
odd

I. If √2 = or
odd
even
(√2) 2 =
squaring both sides, we’ve that
odd
odd
orodd2
even2 odd2
odd2

I. If √2 = or
odd
even
(√2) 2 =
odd
odd
orodd2
even2 odd2
odd2
2 = orodd
even odd
odd
Since odd2 is odd and even2 is even,
so

I. If √2 = or
odd
even
(√2) 2 =
odd
odd
orodd2
even2 odd2
odd2
Clearing the denominator, we've that2 = orodd
even odd
odd
2*even = odd
or 2*odd = odd
so

I. If √2 = or
odd
even
(√2) 2 =
odd
odd
orodd2
even2 odd2
odd2
Clearing the denominator, we've that2 = orodd
even odd
odd
2*even = odd
or 2*odd = odd
so
Both are impossible so √2 can’t be of the form or
odd
even odd .
odd

Next we ague that p/q can’t be of the form
odd
even
.

II. If √2 = p/q =
odd
even
is of the form 2*# (some number), so √2 = odd
2*#
.
odd
even
and any even numbers
.

II. If √2 = p/q =
odd
even
Squaring both sides yields
2 =
odd2
is of the form 2*# (some number), so √2 =
(2*#)(2*#)
odd
2*#
.
.
odd
even
.

II. If √2 = p/q =
odd
even
2 =
odd2
(2*#)(2*#)
odd
2*#
Clearing the denominator, we've that
2*odd2 = (2*#)(2*#)
.
.
odd
even
.

Example C. Evaluate 2x3 – 5x2 + 2x
for x = -2, -1, 3 by factoring it first.
2x3 – 5x2 + 2x = x(2x2 – 5x + 2)
= x(2x – 1)(x – 2)
For x = -2:
(-2)[2(-2) – 1] [(-2) – 2] = -2 [-5] [-4] = -40
For x = -1:
(-1)[2(-1) – 1] [(-1) – 2] = -1 [-3] [-3] = -9
For x = 3:
3 [2(3) – 1] [(3) – 2] = 3 [5] [1] = 15
Applications of Factoring
Your turn: Double check these answers via the expanded form.

II. If √2 = p/q =
odd
even
2 =
odd2
(2*#)(2*#)
odd
2*#
2*odd2 = (2*#)(2*#) This is impossible because
there is one 2 on the left but
only one 2
.
.
odd
even
.

II. If √2 = p/q =
odd
even
2 =
odd2
(2*#)(2*#)
odd
2*#
there’re two or more 2's to the right.only one 2 at least two 2’s
.
.
odd
even
.

II. If √2 = p/q =
odd
even
2 =
odd2
(2*#)(2*#)
odd
2*#
So √2 can’t be a fraction of the form odd .
even
.
.
odd
even
.

II. If √2 = p/q =
odd
even
2 =
odd2
(2*#)(2*#)
odd
2*#
So √2 can’t be a fraction of the form odd .
even
From part I and II, we see √2 fits none of the fractional forms.
Hence √2 is not a fraction.
.
.
odd
even
.

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 D. Find the exact answer.
3
–122. 23. 13
–12527.
8 24. –13
8
3
–2725.
26. –13
64
3
10028. 100029.
3
10,00030. 1,000,00031.
3
0.0132. 0.00133.
3
0.000134. 0.00000135.
3
x

4 radicals and pythagorean theorem x

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