RADICALS
Presented by:
Josaiah Mae Gonzaga
Kerztein Acompañado
9th grade
3
81
Radical Symbol
Radicand
Radical
Index
If there are no
index, it is
assumed to be 2.
Parts of a Radical
Laws of Radicals
Laws of radicals are similar to the laws of integral exponents.
Let a, b, m, n are integers:
Laws Proof
1. The nth root of a product is the product of the nth
roots
𝑛
𝑥𝑦 =𝑛
𝑥 𝑛
𝑦
𝑛
𝑥𝑦 = 𝑥𝑦
1
𝑛 If x is a real number and m is a positive
integer greater than 1, then 𝑚
𝑥 = 𝑥
1
𝑚 = 𝑥
1
𝑛 𝑦
1
𝑛 Power
Rule for Products = 𝑛
𝑥 𝑛
𝑦
2.The nth root of a quotient is the quotient of the nth
roots.
𝑛 𝑥
𝑦
=
𝑛
𝑥
𝑛 𝑦
𝑛 𝑥
𝑦
=
𝑥
𝑦
1
𝑛
If x is a real number and m is a positive
integer greater than 1, then 𝑚
𝑥 = 𝑥
1
𝑚 =
𝑥
1
𝑛
𝑦
1
𝑛
Power Rule
for Quotients =
𝑛
𝑥
𝑛 𝑦
3.
𝑚 𝑛
𝑥 =𝑚𝑛
𝑥 𝑚 𝑛
𝑥 =
𝑚
𝑥
1
𝑛= 𝑥
1
𝑛
1
𝑚 = 𝑥
1
𝑛
1
𝑚 Power of a power
property
= 𝑥
1
𝑛𝑚= 𝑚𝑛
𝑥
4. If x is a real number and n is odd, then 𝑛
𝑥 = 𝑥
5. If x is a real number and n is even, then
𝑛
𝑥𝑛 = 𝑥
Examples:
Apply the laws of radicals:
1. (
3
12)3 = 12
1
3 13
= 12
1
3 = 12
2 . (
5
4) . (5 8) =
5
(4)(8) =
5
32 = 12 , multiply if the indices are the same, just
multiply the radicand, copy the
common index, then simplify.
3.
18
2 =
18
2
divide the radicands if the indices are the same, then simplify
by extracting the square root of 9 The 9 = 3, since 32 = 9
The fifth root of 32 is 2, meaning, when you multiply the root 2 by itself 5 times the answer is 32.
The index 5 indicates the number of times the root be multiplied by itself the radicand. In this to
get case, 25 = 32 Непсе
5
32 = 2
= 9 = 3
Examples:
Apply the laws of radicals:
4.
3
64 = = 2 , the square root of 64 is 8 because the 82 = 64 and
the cube root of 8 = 2, since : 23 = 8.
3
8
5.
8
2 =
8
2
= 4 = 2
divide the radicands if the indices are the same, then simplify
by extracting the square root of 4. The 4 = 2, since 22 = 4
Simplifying
Radical Expressions
Objective:
• Simplify radical expression using the laws of
radicals.
Here are some examples of numbers that are not perfect square:
8 = 4 . 2 = 2 2
12 = 4 . 3 = 2 3
18 = 9 . 2 = 3 2
32 = 16 . 2 = 4 2
40 = 4 . 10 = 2 10
44 = 4 . 11 = 2 11
20 = 4 . 5 = 2 5
24 = 4 . 6 = 2 6
27 = 9 . 3 = 3 3
28 = 4 . 7 = 2 7
45 = 9 . 5 = 3 5
48 = 16 . 3 = 4 3
50 = 25 . 2 = 5 2
63 = 9 . 7 = 3 7
68 = 4 . 17 = 2 17
72 = 36 . 2 = 6 2
52 = 4 . 13 = 2 13
54 = 9 . 6 = 3 6
56 = 4 . 14 = 2 14
60 = 4 . 15 = 2 15
75 = 25 . 3 = 5 3
76 = 4 . 19 = 2 19
80 = 16 . 5 = 4 7
96 = 16 . 6 = 4 6
98 = 49 . 2 = 7 2
99 = 9 . 11 = 3 11
84 = 4 . 21 = 2 21
88 = 4 . 22 = 2 22
90 = 9 . 10 = 3 10
92 = 4 . 23 = 2 23
1.
𝑛
𝑎𝑏 =
𝑛
𝑎𝑏 .
𝑛
𝑎𝑏, the nth root of a product is equal to the nth root of each
factor.
2. 𝑛 𝑚
𝑝 = 𝑚𝑛
𝑝, the nth root of the mth root of a number is equal to the nmth
root of the number.
3.
𝑛 𝑝
𝑞
=
𝑛 𝑝
𝑛 𝑞
, the nth root of the quotient is equal to the nth root of the
numerator and the nth root of the denominator.
The properties of radicals which can be useful in simplifying radical
expressions are as follows:
Simplifying Radical Expressions by:
Reducing the Radicand
01
• Reducing radicand is finding a factor of a radicand whose indicated
roots can be found.
Example:
1. Simplify 50
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
, the factors of 50 are 25 and 2 . 25 is a factor which a
perfect square, meaning you can extract the square root
, the square root of the product is equal to the square root
of the factors.
, the square root of 25 is 5, square root of 2 is already in
simplified form.
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 50 = 5 2 in simplest form
50 is not a perfect square, therefore you have to find a factor of
50 which is a perfect square.
50 = 25. 2
= 25 . 2
= 5 2
2. Simplify
3
81
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: The factor of 81 which is a perfect cube is 27.
3
81
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒
3
81 = 3
3
3 since 3
3
3
3
= 33 3
1
3
3
= 27(3) = 81
=
3
27 .
3
3 = 3
3
3
=
3
27. 3
Another example:
𝑛
𝑎𝑏 = 𝑛
𝑎 .
𝑛
𝑏
1. 8𝑥5𝑦6𝑧13 4.2. 𝑥4. 𝑥. 𝑦6. 𝑧12. 𝑧
=
=
2𝑥2𝑦3𝑧6 2. 𝑥. 𝑧
= 2𝑥2𝑦3𝑧6 2𝑥𝑧
Activity #1.
Simplify the following.
1. 63
2. 99
3.
3
24
4.
3
40
5. 75
1. 63
2. 99
3.
3
24
4.
3
40
5. 75
63 = 9 . 7
= 9 . 7
= 3 7
99 = 9 . 11
= 9 . 11
= 3 11
3
24=
3
8 . 3
=
3
23 .
3
3
= 23
3
=
3
23. 3
Answer key: (2 points each)
3
40=
3
8 . 5
=
3
23 .
3
5
= 23
5
=
3
23. 5
63 = 25 . 3
= 25 . 3
= 5 3
Reducing the order of the
Radicals
02
• To reduce the order of radicals is to reduce the index to its
lowest possible number
Example:
1. Simplify
8
𝑎12
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: , find a factor of radicand that is a power of 8
, find the 8th root of each factor
, reduce the rational exponent to lowest term
8
𝑎12 =
8
𝑎8. 𝑎4
=
8
𝑎8 .
8
𝑎4
= 𝑎
8
𝑎4
= 𝑎 . 𝑎
4
8
= 𝑎 𝑎
, change the radical to exponential form
, change the exponential form to radical form.
= 𝑎 . 𝑎
1
2
Reducing the index to the lowest possible order
𝑚
𝑛
𝑎 = 𝑚𝑛
𝑎 .
𝑛 𝑚
𝑎
Example:
1.
20
32𝑚15𝑛5 =
= 4 5
25 𝑚3 5𝑛5
= 4
2𝑚3𝑛
4 5
25 𝑚3 5𝑛5
Simplifying each radical
Example:
1. 𝑥5 =
=
𝑥4 . 𝑥
=
𝑥2 𝑥
𝑥4 . 𝑥 2.
3
𝑏7 =
3
𝑏6 . 𝑏
3
𝑏6 .
3
𝑏
𝑏2 3
𝑏
=
=
A. Simplify the following.
1. 5
𝑥6 = 5
𝑥5 . 𝑥 = x5
𝑥
2. 36𝑎2 = 36 𝑎2
3.
3
80𝑎10𝑏16 =
3
8. 10𝑎9. 𝑎 . 𝑏15. 𝑏 =
3
23 . 10𝑎9 . 𝑎. 𝑏15. 𝑏
= 62 a = 6𝑎
=
3
23 3
𝑎9 3
𝑏153
10𝑎𝑏 = 2𝑎3𝑏53
10𝑎𝑏
Rationalizing the
Denominator
03
Objective:
• Rationalize a fraction whose denominator
contains a radical.
Rationalizing the denominator of the Radicand
• To rationalize the denominator, we multiply the
denominator by an appropriate expression such
that the product will be a perfect nth root.
• When the denominator is a binomial, we multiply it
by its conjugate. The conjugate of a binomial is a
binomial of the same terms but of different sign.
Simplify:
1)
1
36
2)
3
64 5)
5
4
3)
4
9
4)
1
49
No need to rationalize
because the radical sign is
on the numerator.
=
1
6
=
3
8
=
4
9
=
2
3
=
1
49
=
1
7
=
5
4
=
5
2
Simplify by rationalizing the denominator of the Radicand :
1)
1
7
=
1
7
.
7
7
=
7
7 2
2)
2
𝑥
=
2
𝑥
.
𝑥
𝑥
=
2 𝑥
𝑥 2
3)
2𝑥
𝑦
=
2𝑥
𝑦
.
𝑦
𝑦
=
2𝑥𝑦
𝑦 2
=
7
7
=
2 𝑥
𝑥
=
2𝑥𝑦
𝑦
Simplify by rationalizing the denominators.
1)
5
7
2)
8𝑥
3𝑦
3)
3𝑚
2𝑛
=
5 7
7
=
2 6𝑥𝑦
3𝑦
=
6𝑚𝑛
2𝑛
=
5
7
.
7
7
=
5 7
7 2
=
8𝑥
3𝑦
.
3𝑦
3𝑦
=
24 𝑥𝑦
3𝑦 2 =
4 . 6𝑥𝑦
3𝑦 2
=
3𝑚
2𝑛
.
2𝑛
2𝑛
=
6𝑚𝑛
2𝑛 2
Activity #1
Conjugate of Radical Expressions
Examples:
Radical Expression Conjugate
3 + 2
5 - 2
3 - 2
5 + 2
The product of the radical expression and its conjugate is an integer. In finding its
product is the same procedure in multiplying the sum and difference of 2
binomials. Like, (x + y)(x - y) = x2 - y2
Operations of Radicals
Addition and Subtraction
of Radicals
04
Objective:
• Differentiate like radicals from unlike radicals;
• Add and subtract radical expressions.
Like Radicals
Unlike Radicals
2 3 and -5 3
3 𝑥 and -2 𝑥
5x 2𝑦 and 3y 2𝑦
83
𝑥𝑦 and 3
𝑥𝑦
Same index and same radicand
2 and 6
3
7𝑥 and 7𝑥
4 2 and 4 3 8 and
3
8
Same index and different radicands
Different index and same radicands
Simplify.
1. 3 2 - 5 2
2. 3 𝑎 - 𝑎 + 3 𝑎
3. 6 3𝑎 - 2 𝑎 + 3𝑎 + 5 𝑎
= (3 - 5) 2 = -2 2
= (2 – 1 + 3) 𝑎 = 4 𝑎
= (6 + 1) 3𝑎 + (-2 + 5) 𝑎
= 7 3𝑎 + 3 𝑎
4. 8 + 8 8 + 18 = (1 + 1 + 1) 8 + 18
= 9 2
= 3 8 + 18
= 3 4 . 2 + 9 . 2
= 3 4 2 + 9 2
= 3 . 2 2 +3 2
= 6 2 +3 2
5. 8 + 18 −32 = 4.2 + 9.2 - 16.2
= 2
= 2 2 + 3 2 - 4 2
= (2 + 3 - 4) 2
6. 12𝑎3 - 300𝑎3 = 4𝑎2. 3𝑎 - 100𝑎2. 3𝑎
= 4𝑎2 3𝑎 - 100𝑎2 3𝑎
= 2𝑎 3𝑎 - 10𝑎 3𝑎
= −8a 3𝑎
Find the sum and difference of the following radical expression by matching
column B with column A.
Activity #3: Add, Subtract then Match
Group activity:
Group the following radicals with same radicand, with different
radicand and place them inside each box.
1. 11 3 + 5 3
2. 7 8 + 3 18 - 2 2
3. 5 12 + 5 27 - 4 3
4. 8
2𝑦
7
+ 3
2𝑦
7
-
2𝑦
7
5. 9
3𝑦
25
+
32𝑦
5
- 3
3𝑦
5
Graphs and
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Mercury is the closest planet to the Sun
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Solve the following first-order ordinary differential equation (ODE) using the
method of separation of variables:
dy/dx = 2x + 3
Exercise 2
Consider a mass-spring system described by the second-order ODE:
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and x(t) is the displacement of the mass as a function of time. Solve this ODE
for the displacement x(t) of the mass
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Find the general solution to the second-order linear homogeneous ODE:
d^2y/dx^2 - 4y = 0
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Solve the initial value problem (IVP) for the first-order ODE:
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Simplifying Radical Expressions Mathemat

Simplifying Radical Expressions Mathemat

  • 1.
    RADICALS Presented by: Josaiah MaeGonzaga Kerztein Acompañado 9th grade
  • 2.
    3 81 Radical Symbol Radicand Radical Index If thereare no index, it is assumed to be 2. Parts of a Radical
  • 3.
    Laws of Radicals Lawsof radicals are similar to the laws of integral exponents. Let a, b, m, n are integers:
  • 4.
    Laws Proof 1. Thenth root of a product is the product of the nth roots 𝑛 𝑥𝑦 =𝑛 𝑥 𝑛 𝑦 𝑛 𝑥𝑦 = 𝑥𝑦 1 𝑛 If x is a real number and m is a positive integer greater than 1, then 𝑚 𝑥 = 𝑥 1 𝑚 = 𝑥 1 𝑛 𝑦 1 𝑛 Power Rule for Products = 𝑛 𝑥 𝑛 𝑦 2.The nth root of a quotient is the quotient of the nth roots. 𝑛 𝑥 𝑦 = 𝑛 𝑥 𝑛 𝑦 𝑛 𝑥 𝑦 = 𝑥 𝑦 1 𝑛 If x is a real number and m is a positive integer greater than 1, then 𝑚 𝑥 = 𝑥 1 𝑚 = 𝑥 1 𝑛 𝑦 1 𝑛 Power Rule for Quotients = 𝑛 𝑥 𝑛 𝑦 3. 𝑚 𝑛 𝑥 =𝑚𝑛 𝑥 𝑚 𝑛 𝑥 = 𝑚 𝑥 1 𝑛= 𝑥 1 𝑛 1 𝑚 = 𝑥 1 𝑛 1 𝑚 Power of a power property = 𝑥 1 𝑛𝑚= 𝑚𝑛 𝑥 4. If x is a real number and n is odd, then 𝑛 𝑥 = 𝑥 5. If x is a real number and n is even, then 𝑛 𝑥𝑛 = 𝑥
  • 5.
    Examples: Apply the lawsof radicals: 1. ( 3 12)3 = 12 1 3 13 = 12 1 3 = 12 2 . ( 5 4) . (5 8) = 5 (4)(8) = 5 32 = 12 , multiply if the indices are the same, just multiply the radicand, copy the common index, then simplify. 3. 18 2 = 18 2 divide the radicands if the indices are the same, then simplify by extracting the square root of 9 The 9 = 3, since 32 = 9 The fifth root of 32 is 2, meaning, when you multiply the root 2 by itself 5 times the answer is 32. The index 5 indicates the number of times the root be multiplied by itself the radicand. In this to get case, 25 = 32 Непсе 5 32 = 2 = 9 = 3
  • 6.
    Examples: Apply the lawsof radicals: 4. 3 64 = = 2 , the square root of 64 is 8 because the 82 = 64 and the cube root of 8 = 2, since : 23 = 8. 3 8 5. 8 2 = 8 2 = 4 = 2 divide the radicands if the indices are the same, then simplify by extracting the square root of 4. The 4 = 2, since 22 = 4
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    Objective: • Simplify radicalexpression using the laws of radicals.
  • 9.
    Here are someexamples of numbers that are not perfect square: 8 = 4 . 2 = 2 2 12 = 4 . 3 = 2 3 18 = 9 . 2 = 3 2 32 = 16 . 2 = 4 2 40 = 4 . 10 = 2 10 44 = 4 . 11 = 2 11 20 = 4 . 5 = 2 5 24 = 4 . 6 = 2 6 27 = 9 . 3 = 3 3 28 = 4 . 7 = 2 7 45 = 9 . 5 = 3 5 48 = 16 . 3 = 4 3 50 = 25 . 2 = 5 2 63 = 9 . 7 = 3 7 68 = 4 . 17 = 2 17 72 = 36 . 2 = 6 2 52 = 4 . 13 = 2 13 54 = 9 . 6 = 3 6 56 = 4 . 14 = 2 14 60 = 4 . 15 = 2 15 75 = 25 . 3 = 5 3 76 = 4 . 19 = 2 19 80 = 16 . 5 = 4 7 96 = 16 . 6 = 4 6 98 = 49 . 2 = 7 2 99 = 9 . 11 = 3 11 84 = 4 . 21 = 2 21 88 = 4 . 22 = 2 22 90 = 9 . 10 = 3 10 92 = 4 . 23 = 2 23
  • 10.
    1. 𝑛 𝑎𝑏 = 𝑛 𝑎𝑏 . 𝑛 𝑎𝑏,the nth root of a product is equal to the nth root of each factor. 2. 𝑛 𝑚 𝑝 = 𝑚𝑛 𝑝, the nth root of the mth root of a number is equal to the nmth root of the number. 3. 𝑛 𝑝 𝑞 = 𝑛 𝑝 𝑛 𝑞 , the nth root of the quotient is equal to the nth root of the numerator and the nth root of the denominator. The properties of radicals which can be useful in simplifying radical expressions are as follows:
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    Reducing the Radicand 01 •Reducing radicand is finding a factor of a radicand whose indicated roots can be found.
  • 13.
    Example: 1. Simplify 50 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: ,the factors of 50 are 25 and 2 . 25 is a factor which a perfect square, meaning you can extract the square root , the square root of the product is equal to the square root of the factors. , the square root of 25 is 5, square root of 2 is already in simplified form. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 50 = 5 2 in simplest form 50 is not a perfect square, therefore you have to find a factor of 50 which is a perfect square. 50 = 25. 2 = 25 . 2 = 5 2
  • 14.
    2. Simplify 3 81 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: Thefactor of 81 which is a perfect cube is 27. 3 81 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 3 81 = 3 3 3 since 3 3 3 3 = 33 3 1 3 3 = 27(3) = 81 = 3 27 . 3 3 = 3 3 3 = 3 27. 3
  • 15.
    Another example: 𝑛 𝑎𝑏 =𝑛 𝑎 . 𝑛 𝑏 1. 8𝑥5𝑦6𝑧13 4.2. 𝑥4. 𝑥. 𝑦6. 𝑧12. 𝑧 = = 2𝑥2𝑦3𝑧6 2. 𝑥. 𝑧 = 2𝑥2𝑦3𝑧6 2𝑥𝑧
  • 16.
    Activity #1. Simplify thefollowing. 1. 63 2. 99 3. 3 24 4. 3 40 5. 75
  • 17.
    1. 63 2. 99 3. 3 24 4. 3 40 5.75 63 = 9 . 7 = 9 . 7 = 3 7 99 = 9 . 11 = 9 . 11 = 3 11 3 24= 3 8 . 3 = 3 23 . 3 3 = 23 3 = 3 23. 3 Answer key: (2 points each) 3 40= 3 8 . 5 = 3 23 . 3 5 = 23 5 = 3 23. 5 63 = 25 . 3 = 25 . 3 = 5 3
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    Reducing the orderof the Radicals 02 • To reduce the order of radicals is to reduce the index to its lowest possible number
  • 19.
    Example: 1. Simplify 8 𝑎12 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: ,find a factor of radicand that is a power of 8 , find the 8th root of each factor , reduce the rational exponent to lowest term 8 𝑎12 = 8 𝑎8. 𝑎4 = 8 𝑎8 . 8 𝑎4 = 𝑎 8 𝑎4 = 𝑎 . 𝑎 4 8 = 𝑎 𝑎 , change the radical to exponential form , change the exponential form to radical form. = 𝑎 . 𝑎 1 2
  • 20.
    Reducing the indexto the lowest possible order 𝑚 𝑛 𝑎 = 𝑚𝑛 𝑎 . 𝑛 𝑚 𝑎 Example: 1. 20 32𝑚15𝑛5 = = 4 5 25 𝑚3 5𝑛5 = 4 2𝑚3𝑛 4 5 25 𝑚3 5𝑛5
  • 21.
    Simplifying each radical Example: 1.𝑥5 = = 𝑥4 . 𝑥 = 𝑥2 𝑥 𝑥4 . 𝑥 2. 3 𝑏7 = 3 𝑏6 . 𝑏 3 𝑏6 . 3 𝑏 𝑏2 3 𝑏 = =
  • 22.
    A. Simplify thefollowing. 1. 5 𝑥6 = 5 𝑥5 . 𝑥 = x5 𝑥 2. 36𝑎2 = 36 𝑎2 3. 3 80𝑎10𝑏16 = 3 8. 10𝑎9. 𝑎 . 𝑏15. 𝑏 = 3 23 . 10𝑎9 . 𝑎. 𝑏15. 𝑏 = 62 a = 6𝑎 = 3 23 3 𝑎9 3 𝑏153 10𝑎𝑏 = 2𝑎3𝑏53 10𝑎𝑏
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    Objective: • Rationalize afraction whose denominator contains a radical.
  • 25.
    Rationalizing the denominatorof the Radicand • To rationalize the denominator, we multiply the denominator by an appropriate expression such that the product will be a perfect nth root. • When the denominator is a binomial, we multiply it by its conjugate. The conjugate of a binomial is a binomial of the same terms but of different sign.
  • 26.
    Simplify: 1) 1 36 2) 3 64 5) 5 4 3) 4 9 4) 1 49 No needto rationalize because the radical sign is on the numerator. = 1 6 = 3 8 = 4 9 = 2 3 = 1 49 = 1 7 = 5 4 = 5 2
  • 27.
    Simplify by rationalizingthe denominator of the Radicand : 1) 1 7 = 1 7 . 7 7 = 7 7 2 2) 2 𝑥 = 2 𝑥 . 𝑥 𝑥 = 2 𝑥 𝑥 2 3) 2𝑥 𝑦 = 2𝑥 𝑦 . 𝑦 𝑦 = 2𝑥𝑦 𝑦 2 = 7 7 = 2 𝑥 𝑥 = 2𝑥𝑦 𝑦
  • 28.
    Simplify by rationalizingthe denominators. 1) 5 7 2) 8𝑥 3𝑦 3) 3𝑚 2𝑛 = 5 7 7 = 2 6𝑥𝑦 3𝑦 = 6𝑚𝑛 2𝑛 = 5 7 . 7 7 = 5 7 7 2 = 8𝑥 3𝑦 . 3𝑦 3𝑦 = 24 𝑥𝑦 3𝑦 2 = 4 . 6𝑥𝑦 3𝑦 2 = 3𝑚 2𝑛 . 2𝑛 2𝑛 = 6𝑚𝑛 2𝑛 2 Activity #1
  • 29.
    Conjugate of RadicalExpressions Examples: Radical Expression Conjugate 3 + 2 5 - 2 3 - 2 5 + 2 The product of the radical expression and its conjugate is an integer. In finding its product is the same procedure in multiplying the sum and difference of 2 binomials. Like, (x + y)(x - y) = x2 - y2
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    Objective: • Differentiate likeradicals from unlike radicals; • Add and subtract radical expressions.
  • 33.
    Like Radicals Unlike Radicals 23 and -5 3 3 𝑥 and -2 𝑥 5x 2𝑦 and 3y 2𝑦 83 𝑥𝑦 and 3 𝑥𝑦 Same index and same radicand 2 and 6 3 7𝑥 and 7𝑥 4 2 and 4 3 8 and 3 8 Same index and different radicands Different index and same radicands
  • 34.
    Simplify. 1. 3 2- 5 2 2. 3 𝑎 - 𝑎 + 3 𝑎 3. 6 3𝑎 - 2 𝑎 + 3𝑎 + 5 𝑎 = (3 - 5) 2 = -2 2 = (2 – 1 + 3) 𝑎 = 4 𝑎 = (6 + 1) 3𝑎 + (-2 + 5) 𝑎 = 7 3𝑎 + 3 𝑎
  • 35.
    4. 8 +8 8 + 18 = (1 + 1 + 1) 8 + 18 = 9 2 = 3 8 + 18 = 3 4 . 2 + 9 . 2 = 3 4 2 + 9 2 = 3 . 2 2 +3 2 = 6 2 +3 2
  • 36.
    5. 8 +18 −32 = 4.2 + 9.2 - 16.2 = 2 = 2 2 + 3 2 - 4 2 = (2 + 3 - 4) 2 6. 12𝑎3 - 300𝑎3 = 4𝑎2. 3𝑎 - 100𝑎2. 3𝑎 = 4𝑎2 3𝑎 - 100𝑎2 3𝑎 = 2𝑎 3𝑎 - 10𝑎 3𝑎 = −8a 3𝑎
  • 37.
    Find the sumand difference of the following radical expression by matching column B with column A. Activity #3: Add, Subtract then Match
  • 38.
    Group activity: Group thefollowing radicals with same radicand, with different radicand and place them inside each box. 1. 11 3 + 5 3 2. 7 8 + 3 18 - 2 2 3. 5 12 + 5 27 - 4 3 4. 8 2𝑦 7 + 3 2𝑦 7 - 2𝑦 7 5. 9 3𝑦 25 + 32𝑦 5 - 3 3𝑦 5
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    Differential equations Mercury isthe closest planet to the Sun and the smallest one in the Solar System—it’s only a bit larger than the Moon Venus has a beautiful name and is the second planet from the Sun. It’s hot and has a poisonous atmosphere Basics Fundamentals
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    Our schools School 1 Mercuryis the closest planet to the Sun School 2 Venus is the second planet from the Sun School 3 Earth is the third planet from the Sun
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    Exercise 1 Solve thefollowing first-order ordinary differential equation (ODE) using the method of separation of variables: dy/dx = 2x + 3
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    Exercise 2 Consider amass-spring system described by the second-order ODE: m(d^2x/dt^2) + c(dx/dt) + kx = 0 where m is the mass, c is the damping coefficient, k is the spring constant, and x(t) is the displacement of the mass as a function of time. Solve this ODE for the displacement x(t) of the mass
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    Exercise 3 Find thegeneral solution to the second-order linear homogeneous ODE: d^2y/dx^2 - 4y = 0
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    Exercise 4 Solve theinitial value problem (IVP) for the first-order ODE: dy/dx = -2y, with the initial condition y(0) = 5
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Editor's Notes

  • #21 Other example: 3//64 = 3/8 = 2 
  • #22 If it’s not divisible by 2 or the given index, we should reduce the radicand