When given the lenghts of all 3 sides of a triangle, how do we determine if the triangle is right, acute or obtuse?
Recall that for a right triangle with sides of length a,b and c (with c being the hypotenuse),
a^2 + b^2 = c^2
So a triangle with sides 7, 24 and 25 is a right triangle
7^2 + 24^2 = 25^2
49 + 576 = 625
If a^2 + b^2 < c^2, the triangle is obtuse.
A triangle with sides 4, 6 and 10 is an example of an obtuse triangle
4^2 + 6^2 < 10^2
16 + 36 < 100
52 < 100
If a^2 + b^2 > c^2, the triangle is acute.
A triangle with sides 7, 8, 9 is an acute triangle
7^2 + 8^2 > 9^2
49 + 64 > 81
113 > 81
Wednesday, May 30, 2012
Saturday, May 26, 2012
Suppose you wish to find the asymptotes of the following:
2x/(3x + 2)
To find the vertical asymptote, find where the denominator is undefined for x. Set the denominator equal to zero.
3x + 2 = 0
x = 2/3
Therefore the vertical asymptote is the line x = 2/3
To find the horizontal asymptote, divide the coefficients of the x term (numerator/denominator). Notice the coefficients are 2 and 3.
The horizontal aymptote is the line y = 2/3.
The graph will approach but never touch the asymptotes.
2x/(3x + 2)
To find the vertical asymptote, find where the denominator is undefined for x. Set the denominator equal to zero.
3x + 2 = 0
x = 2/3
Therefore the vertical asymptote is the line x = 2/3
To find the horizontal asymptote, divide the coefficients of the x term (numerator/denominator). Notice the coefficients are 2 and 3.
The horizontal aymptote is the line y = 2/3.
The graph will approach but never touch the asymptotes.
Thursday, May 24, 2012
Some basic trigonometric identities are as follows:
sin^2(x) + cos^2(x) = 1
1 + tan^2(x) = sec^2(x)
1 + cot^2(x) = csc^2(x)
Recall that
sin(x) = 1/csc(x)
cos(x) = 1/sec(x)
tan(x) = 1/cot(x)
The trigonometric functions and identities and can derived from the basic trigonometric functions sin(x) and cos(x).
For example:
sin(x) = opposite/hypotenuse, cos(x) = adjacent/hypotenuse, tan(x) = opposite/adjacent,
therefore tan(x) = sin(x)/cos(x)
1 + tan^2(x) = sec^2(x) using sin(x) and cos(x)
1+ sin^2(x)/cos^2(x) = 1/cos^2(x)
[cos^2(x) + sin^2(x)]/cos^2(x) = 1/cos^2(x)
multiply both sides by cos^2(x) to get
cos^2(x) + sin^2(x) = 1, which confirms the first identity.
sin^2(x) + cos^2(x) = 1
1 + tan^2(x) = sec^2(x)
1 + cot^2(x) = csc^2(x)
Recall that
sin(x) = 1/csc(x)
cos(x) = 1/sec(x)
tan(x) = 1/cot(x)
The trigonometric functions and identities and can derived from the basic trigonometric functions sin(x) and cos(x).
For example:
sin(x) = opposite/hypotenuse, cos(x) = adjacent/hypotenuse, tan(x) = opposite/adjacent,
therefore tan(x) = sin(x)/cos(x)
1 + tan^2(x) = sec^2(x) using sin(x) and cos(x)
1+ sin^2(x)/cos^2(x) = 1/cos^2(x)
[cos^2(x) + sin^2(x)]/cos^2(x) = 1/cos^2(x)
multiply both sides by cos^2(x) to get
cos^2(x) + sin^2(x) = 1, which confirms the first identity.
Thursday, May 17, 2012
Suppose we wish to solve the following problem:
2x/(x + 1) + 3/x = 12
First we need to get a common denominator. Notice the factors in the denominators are (x+1) for the first term and x for the second term. We multiply the factors to get x(x+ 1) for the common denominator.
The numerator and denominator of the first term is multiplied by x and the numerator and denominator of the second term is multiplied by (x + 1). The 12 is multiplied by x(x + 1)/x(x + 1).
This gives us
2x(x)/x(x+1) + 3(x+ 1)/x(x + 1) = 12x(x + 1)/x(x + 1)
When solving this, we can ignore the denominators. The resason for this is since all the denominators are the same, we can multiply the entire equation by x(x + 1), which cancels the denominators
This leaves us with
2x(x) + 3(x + 1) = 12x(x + 1)
2x^2 + 3x + 3 = 12x^2 + 12x
0 = 10x^2 + 9x  3
Using quadratic formula we get
x = [(9 +/ sqrt(81 4(10)(3))]/2(10)
x = [9 +/ sqrt(201)]/20
Solving for x, we get x is approximately 0.259 or 1.16
2x/(x + 1) + 3/x = 12
First we need to get a common denominator. Notice the factors in the denominators are (x+1) for the first term and x for the second term. We multiply the factors to get x(x+ 1) for the common denominator.
The numerator and denominator of the first term is multiplied by x and the numerator and denominator of the second term is multiplied by (x + 1). The 12 is multiplied by x(x + 1)/x(x + 1).
This gives us
2x(x)/x(x+1) + 3(x+ 1)/x(x + 1) = 12x(x + 1)/x(x + 1)
When solving this, we can ignore the denominators. The resason for this is since all the denominators are the same, we can multiply the entire equation by x(x + 1), which cancels the denominators
This leaves us with
2x(x) + 3(x + 1) = 12x(x + 1)
2x^2 + 3x + 3 = 12x^2 + 12x
0 = 10x^2 + 9x  3
Using quadratic formula we get
x = [(9 +/ sqrt(81 4(10)(3))]/2(10)
x = [9 +/ sqrt(201)]/20
Solving for x, we get x is approximately 0.259 or 1.16
Monday, May 14, 2012
Great time to get the books before final exam, especially now, on the lulu homepage, there will be a name of a savings coupon that they can add your cart to save. This promotion ends Friday May, 18 at 11:59 PM. you must go to the homepage www.lulu.com, but my books are at http://www.lulu.com/spotlight/KKauffman1969
Here's part of the second book, chapter on sequences and series.
Here's part of the second book, chapter on sequences and series.
Chapter 7
Sequences and Geometric Applications
When
we think of a sequence, we think of an ordered list. For example, a
history teacher might discuss a sequence of battles during World War
II. A baseball historian might discuss the sequence of events
leading up to the first organized professional baseball league. In
terms of mathematics, a sequence
refers to a list of numbers written in a specific order.
Sequences
can either be finite, which contains a set number of terms, or
infinite, where the sequence continues on forever. A finite sequence
is a function whose domain is the set of all natural numbers for a
specified number n.
An infinite sequence is a function whose domain is the set of all
natural numbers. An easier way to denote terms of a sequence is to
use a_{n}
,
which denotes the nth
term of the sequence. Similarly a_{1}
denotes the first term in the sequence, a_{2
}denotes
the second term of the sequence and so on.
Some
examples of finite and infinite sequences are as follows:
Finite:
2, 5, 8, 11, 14, 17, 20, 23, 26
1,
1, 3, 5, 7, 9, 11, 13, 15
Infinite:
1, 2, 3, 4, 5, 6, 7, …....
4,
8, 12, 16, 20, 24, 28, …...
Consider
the following infinite sequence:
2,
8, 32, 128, 512, …..
We
can define the nth
term of the sequence by finding the relationship between the terms.
Notice a_{1}
= 2 and a_{2}
is 4 times a_{1}.
It is also 6 more than a_{1}
but that is not the pattern throughout the sequence. If it was the
pattern, a_{3}
would be 8 + 6 = 14, but the third term is 32, which is 4 times 8.
The formula holds true for the next term as well since 32 times 4
equals 128. Therefore the formula for the nth
term in the sequence is a_{n}
= 4a_{n}_{1}
When
we apply the formula we just got for a_{n} , we
obtain the first 5 numbers in the series and therefore we know our
formula is correct.
a_{1
}=
2
a_{2}
= 4(a_{1})
= 4(2) = 8
a_{3}
= 4(a_{2})
= 4(8) = 32
a_{4}
= 4(a_{3})
= 4(32) = 128
a_{5}
= 4(a_{4})
= 4(128) = 512
Example:
Find
the first 6 terms and the 46^{th}
term of the infinite sequence with a_{n}
= 5n
– 11.
To
find the first 6 terms of the sequence, we need to substitute 1, 2,
3, 4, 5 and 6 for n
in the formula. Recall that the first 6 terms of the sequence are
denoted as a_{1},
a_{2},
a_{3},
a_{4},
a_{5
}and
a_{6}.
Therefore,
a_{1}
= 5(1) – 11 = 5 – 11 = 6
a_{2}
= 5(2) – 11 = 10 – 11 = 1
a_{3}
= 5(3) – 11 = 15 – 11 = 4
a_{4}
= 5(4) – 11 = 20 – 11 = 9
a_{5}
= 5(5) – 11 = 25 – 11 = 14
a_{6}
= 5(6) – 11 = 30 – 11 = 19
Therefore,
the first 6 terms of the sequence are 6, 1, 4, 9, 14, 19
• Note
the pattern is each term is 5 greater than the previous term. Once
that pattern has been established, we can just add 5 to the previous
term and don't have to perform each calculation. In the next part
where we have to find the 46^{th}
term, it's easier to perform the calculation instead of adding 5 each
time until we reach the 46^{th}
term.
To
find the 46^{th}
term of the sequence, we need to substitute 46 for n in
the formula.
Therefore,
a_{46}
= 5(46) – 11 = 230 11 = 219.
Arithmetic Sequence
In
another example, suppose the sequence is 2, 8, 14, 20, 26, 32, 38,
….... Notice that each term is 6 more than the previous term. A
sequence such as this is known as an
arithmetic sequence because
each term is found by adding the same number to the previous term.
An arithmetic sequence is a sequence in the form
a_{1.},
a_{1}
+ d,
a_{1}
+ 2d,
a_{1}
+ 3d,
a_{1}
+ 4d,
…... a_{1}
+ (n
– 1)d,
where a_{1}
is the first term of the sequence and d
is the
common difference between
the terms.
The
nth term
of an arithmetic sequence is given by the formula a_{n}
= a_{1}
+ (n – 1)d.
Consider
the following arithmetic sequences below. Notice the common
difference by subtracting the first term from the second, the second
term from the third and so on.
Finite
arithmetic sequence: 3, 10, 17, 24, 31, 38
The
common difference d is 7 since 10 – 3 = 7, 17 – 10 = 7,
24 – 17 = 7, 31 – 24 = 7 and 38 – 31 = 7.
Infinite
arithmetic sequence: 6, 3, 0, 3, 6, 9, ….....
The
common difference d is 3 since 3  6 = 3, 3  0 = 3, 6 
(3) = 3, 9  (6) = 3 and so on.
Example:
In an arithmetic sequence, the first term is 8 and the common
difference is 2. What are the first 6 terms of the sequence and
what is the 29^{th} term?
To
solve this problem we start with 8 and add the common difference to
each term thereafter until we get all 6 terms.
Therefore,
a_{1}
= 8, a_{2}
= 8 – 2 = 6, a_{3
}= 6 – 2 = 4, a_{4}
= 4 – 2 = 2, a_{5}
= 2 – 2 = 0, a_{6
}= 0 – 2 = 2
The
first 6 terms in the series are 8, 6, 4, 2, 0, 2.
To
find the 29^{th}
term in the sequence, we use the formula to find the nth
term in a sequence. Substitute
29 for n in the
formula a_{n}_{
}= a_{1}
+ (n – 1)d
to get
a_{29
}= 8 + (29 – 1)(2)
a_{29}
= 8 + (28)(2)
a_{29}
= 8 – 56
a_{29}
= 48
Example:
The first 4 terms of an
arithmetic sequence are 4, 10, 16 and 22. What is the 95^{th}
term?
To
solve this problem we have to find the common difference between each
term. Since a_{1}
= 4 and a_{2}
= 10, the common
difference is 10 – 4 = 6.
Now
we use the formula a_{n}_{
}= a_{1}
+ (n
– 1)d and
substitute 95 for n
and 6 for d
and solve for a_{n
}to get
a_{95}
= 4 + (95 – 1)(6)
a_{95}
= 4 + 94(6)
a_{95}
= 4 + 564
a_{95}^{
}=
568
Sometimes
we don't know the first several terms of a sequence. Maybe we know
the first term and the 30^{th}
term and we need to find the 15^{th}
term. How do we approach such a problem?
First,
we have to determine the common difference by substituting the value
of the 30^{th}
term for an, the value of the first term for a_{1
}and
30 for n
into the formula a_{n}_{
}=
a_{1}
+ (n
– 1)d.
Example:
The first term of an
arithmetic sequence is 6 and the 45^{th}
term is 490. What are the first 5 terms of the sequence?
To
solve this problem we need to find the common difference. We use the
formula a_{n}_{
}= a_{1}
+ (n
– 1)d. Since
the 45^{th}
term is 490, we substitute 490 for a_{n}_{
}. Since the first
term is 6, we substitute 6 for a_{1}
and 45 for n to
get
490
= 6 + (45 – 1)d
490
= 6 + 44d
484
= 44d
11
= d
Since
the common difference is 11, the first 5 terms of the sequence are 6,
17, 28, 39, 50 and are determined as follows
a_{1}
= 6
a_{2}
= 6 + 11 = 17
a_{3}
= 17 + 11 = 28
a_{4}
= 28 + 11 = 39
a_{5}
= 39 + 11 = 50
___________________________________________________________________________________________________
Suppose
we know two terms of an arithmetic sequence and we wish to insert
numbers between them. The numbers are known as arithmetic means.
If a single number is inserted between the two numbers, it's known
as the arithmetic mean between the two numbers.
For
example, the numbers 4 and 7 are the arithmetic means between 1 and
10 in the sequence 1, 4, 7, 10. The common difference between the
terms is 3. The number 6 is the arithmetic mean between 3 and 9 in
the sequence 3, 6, 9. The common difference between the terms is 3.
How
do we determine the arithmetic means? Consider the following
examples.
Example:
Find two arithmetic means
between 8 and 23.
To
solve this problem, we have to determine how many terms there will be
and then find the common difference between the terms.
We
know a_{1
}= 8 but we don't know
a_{2}
or a_{3}.
We know that a_{4}
= 23. There is a jump from a_{1}
to a_{2},
from a_{2
}to a_{3}
and from a_{3}
to a_{4}.
Therefore there are 3 equal distances going from 8 to 23. The total
distance from 8 to 23 is 15 since 23 – 8 = 15. Now we take 15
divided by 3 to get 5. Therefore the common difference is 5. The
arithmetic means, a_{2}
and a_{3}
are then
a_{2}
= a_{1}
+ 5 = 8 + 5 = 13, a_{3}
= a_{2
}+
5 = 13 + 5 = 18
Friday, May 11, 2012
Time is running out before those final exams. Any kids having algebra trouble will benefit from my books. The book cost is cheaper than a 1 hour tutoring session. http://www.lulu.com/spotlight/KKauffman1969
Wednesday, May 9, 2012
Here's part of chapter 2 from my second book, Algebra Simplified Intermediate & Advanced. If interested in the book, go to http://www.lulu.com/spotlight/KKauffman1969
1^{2} = 1 10^{2} = 100 19^{2} = 361
729/9 = 81 and we know the square root of 81 is 9.
(3x)^{3}
= 3x
∙ 3x
∙ 3x
= 27x^{3}.
Example: Simply the following.
13. √(80x^{5}) 14. √(75xy^{2}z^{3}) 15. √(98z^{3})
Or we can think of 27^{2/3 }as
Chapter 2
Radicals, Rational Exponents and Complex Numbers
Square Roots, Cube Roots and Higher Roots
Recall
when raising a number to a power n,
where n
is an integer greater than 1, we multiply the number by itself n
times. For example, 4^{3
}=
4 ∙ 4 ∙ 4. Now suppose we want to know what number multiplied by
itself 2 times equals 169. Problems of this kind can be represented
using radicals.
A
radical
symbol √ is
used to show the square
root,
or
principal square root
of a number or expression that appears under the radical symbol.
Recall that the square root is defined as a number or expression
multiplied by itself twice to equal the number or expression under
the radical symbol, known as the
radicand.
For
example, if we want to know what number multiplied by itself 2 times
equals 169, we can set this up with the radical symbol as follows:
√169,
read as “square root of 169”. The answer to this is 13.

√169, read as “negative square root of 169”. The answer to
this is 13.
√0.09
= 0.3 and 0.3 since (0.3)^{2}
and (0.3)^{2}
equals 0.09. The principal square root is 0.3. Another way to
simplify this is to change √0.09 to √(9/100) and simplify to
3/10.
√(25/49)
= 5/7 and 5/7 since (5/7)^{2}
and (5/7)^{2}.
The principal square root is 5/7.
• Note
that a square root also has a negative value since a negative times a
negative equals a positive, but we will deal with only the principal
square root unless otherwise noted.
• Note
that you can also simplify the square root of a fraction by taking
the square root of the numerator and then the square root of the
denominator instead of the square root of the fraction as a whole.
In the previous example, you can take the square root of 25 first,
then the square root of 49.
• Note
that the square root of many positive integers are not whole numbers
or rational numbers. For example, √19 can be found on a calculator
or by leaving the answer as √19.
An
easy way to solve many square root problems is to know the perfect
squares from 1 to 25. They are as follows:
1^{2} = 1 10^{2} = 100 19^{2} = 361
2^{2}
= 4 11^{2}
= 121 20^{2}
= 400
3^{2}
= 9 12^{2}
= 144 21^{2}
= 441
4^{2}
= 16 13^{2}
= 169 22^{2}
= 484
5^{2}
= 25 14^{2}
= 196 23^{2}
= 529
6^{2}
= 36 15^{2}
= 225 24^{2}
= 576
7^{2}
= 49 16^{2}
= 256 25^{2}
= 625.
8^{2}
= 64 17^{2}
= 289
9^{2}
= 81 18^{2}
= 324
Sometimes
we have to find the square root of a number that is not a perfect
square. In these cases, we break down the radicand into factors, one
of which is a perfect square.
Examples:
Find each square root.
1.
√68
First,
find factors of 68.
Since
68 is even, we can divide it by 2. Therefore, 68 = 2 ∙ 34. Notice
34 is also divisible by 2, therefore 34 = 2 ∙ 17.
So
68 is factored into 2 ∙ 2 ∙ 17. Notice that 17 is prime and
cannot be factored further and 2 ∙ 2 = 4, which is a perfect
square. Therefore √68 = √4 ∙ √17 = 2√17.
2.
√108
First,
find factors of 108.
Since
108 is even, we can divide it by 2. Therefore 108 = 2 ∙ 54. Notice
54 is also divisible by 2, therefore 54 = 2 ∙ 27. Next, we know
that 27 = 3 ∙ 3 ∙ 3.
The
factors of 108 are 2 ∙ 2 ∙ 3 ∙ 3 ∙ 3. Notice 2 ∙ 2 = 4,
which is a perfect square and 3 ∙ 3 = 9, which is also a perfect
square. Therefore, 108 = 4 ∙ 9 ∙ 3 and √108 = √4 ∙ √9 ∙
√3 = 2 ∙ 3 ∙ √3 = 6√3.
• Note
that 108 = 36 ∙ 3 and 36 is a perfect square. But if you can't see
right away that 3 is a factor of 108, you can break down by dividing
108 by 2 first and then simplify further at that point. It's easy to
determine that 108 is divisible by 3. If the sum of the digits of a
number are divisible by 3, the number is divisible by 3.
Recall,
for a variable x, where x ≠ 0, √x^{2}
= x and x. If we are dealing with just the principal
square root, then the answer is x. Recall that x
is the absolute value of x.
Examples:
Find the square root of the
following.
a.
√(25x^{2})
Simplify
the numerical part and the variable part separately.
√25
= 5 since 5^{2}
= 25 and √x^{2}
= x
since (x)^{2}
= x^{2}.
Therefore,
√(25x^{2})
= 5x.
b.
√(64x^{4}y^{6})
√64
= 8 since 8^{2}
= 64, √x^{4}
= x^{2}
since (x^{2})^{2}
= x^{4}
and √y^{6}
= y^{3}
since (y^{3})^{2}
= y^{6}.
Therefore,
multiply the terms in bold to get √(64x^{4}y^{6})
= 8x^{2}y^{3}.
c.
√(x
+ 4)^{2
}
= (x
+
4) because (x
+ 4)(x
+
4) = (x
+ 4)^{2}
• Note
that the square root of any number or expression squared is just that
number. For example, notice in the previous examples that √(x
+ 4)^{2 } = (x + 4) and √25 = √(5)^{2} =
5.
d.
√(x^{2}
+ 12x
+ 36) = √(x
+ 6)(x
+ 6)
Notice
that we factored (x^{2}
+ 12x
+ 36) to get (x
+ 6)(x
+ 6). That is a perfect square which enables us to simplify the
radical expression.
√(x^{2}
+ 12x
+ 36) = √(x
+ 6)(x
+ 6) = √(x
+
6)^{2}
= (x
+ 6)
• Note
that you can check by squaring your answer. After squaring, you
should get the expression under the √.
In
the previous examples, we simplified radicals that were perfect
squares. Many times we can simplify radicals that are not perfect
squares. The idea is to break down the number or variable into
factors, one of which is a perfect square. By the
multiplication
property of radicals ^{n}√(ab)
= ^{n}√a
∙ ^{n}√b,
where ^{n}√a
and ^{n}√b
are real numbers.
• Note
that in a square root n is 2 but is not written. When solving
any root higher than 2, it is noted in the upper left of the radical.
For example, the cube root (n = 3) of a is noted as
^{3}√a.
The
cube
root of
a number n
is the number when multiplied by itself 3 times equals n.
For example, the cube root of 125 is 5 because (5) ∙ (5) ∙
(5 ) = 125, or (5)^{3}
= 125. By
definition, the cube root of a is defined as ^{3}√a
= b,
if b^{3}
= a.
Examples:
^{3}√27
= 3 since 3^{3}
= 27
^{3}√(27)
= 3 since (3)^{3}
= 27
^{3}√125
= 5 since 5^{3}
= 125
^{3}√(125)
= 5 since (5)^{3}
= 125
• Note
that 27 and 125 have 1 real number cube root and 2 real number square
roots. Any negative number will have a cube root that is negative
and any positive number will have a cube root that is positive.
Examples:
Simplify the following.
a.
^{3}√8
To
solve this, think of what number multiplied by itself 3 times gives
you 8. Notice that (2) ∙ (2) ∙ (2) = 8. Therefore, ^{3}√8
= 2.
b.
^{3}√343
= 7 since 7 ∙ 7 ∙ 7 = 343.
c.
^{3}√729
= 9
• Note
that if the sum of the digits of a number is divisible by 9, the
number is divisible by 9.
The
sum of the digits of 729 is 18, which is divisible by 9. Therefore,
729 is divisible by 9 and 729 is divisible by 9.
729/9 = 81 and we know the square root of 81 is 9.
We
reviewed how to take the cube root of positive and negative numbers.
Now we will review how to take the cube root involving variables.
Recall that by definition, if x
is a variable representing a real number, ^{3}√x^{3}
= x.
Examples:
Simplify the following.
a.
^{3}√(27x^{3})
= ^{3}√(27)
∙ ^{3}√(x^{3})
= 3 ∙ x
= 3x
Check
by taking the cube of 3x.
b.
^{3}√(5x
+ 2y)^{3}
= (5x
+ 2y)
because of the rule that ^{3}√x^{3}
= x.
c.
^{3}√(216x^{6}y^{12})
^{3}√(216)
= 6, ^{3}√(x^{6})
= x^{2 }^{
}since (x^{2})^{3}
= x^{6
}and ^{3}√(y^{12})
= y^{4}
since (y^{4})^{3
}= y^{12}.
^{ }Multiply
the terms in bold to get 6x^{2}y^{4}.
The
previous examples were perfect cubes, but oftentimes we can find the
cube root of expressions that are not perfect cubes. The idea is to
break down the number or variable into factors, one of which is a
perfect cube. For example, if you solve ^{3}√81,
break it down into ^{3}√27
∙ ^{3}√3
because ^{3}√27
= 3. If you solve ^{3}√(y^{5}),
break it down into ^{3}√(y^{3})
∙ ^{3}√(y^{2})
because ^{3}√(y^{3})
= y.
Remember from the multiplication property of radicals that ^{3}√ab
= ^{3}√a
∙ ^{3}√b.
Example:
Simplify the following.
^{3}√(16x^{4})
^{3}√16
= ^{3}√8
∙ ^{3}√2
= 2
∙ ^{3}√2
(^{3}√8
= 2 because 2^{3}
= 8)
^{3}√(x^{4})
= ^{3}√(x^{3})
∙ ^{3}√x
= x
∙ ^{3}√x
(^{3}√(x^{3})
= x by
the definition of cube root (x)^{3}
= x^{3})
Multiply
like terms to get ^{3}√(16x^{4}) =
2x∙^{3}√(2x)
• Note
that you can check by cubing the answer, [2x ∙ ^{3}√(2x)]^{3
}= 8x^{3} ∙ 2x = 16x^{4}.
Example: Simply the following.
^{3}√(192x^{4}y^{5})
^{3}√(192)
= ^{3}√(64)
∙ ^{3}√(3)
= 4∙
^{3}√(3)
(note that 64 ∙ 3 = 192 and (4)^{3}
= 64)
^{3}√(x^{4})
= ^{3}√(x^{3})
∙ ^{3}√(x)
= x
∙ ^{3}√(x)
(note that x^{3}∙
x
= x^{4}
and (x)^{3}
= x^{3})
^{3}√(y^{5})
= ^{3}√(y^{3})
∙ ^{3}√(y^{2})
= y
∙ ^{3}√(y^{2})
(note that y^{3}
∙ y^{2}
= y^{5}
and (y)^{3}
= y^{3})
Now
multiply the like terms to get 4xy∙^{3}√(3xy^{2}).
In
some cases cube roots will be the form of a fraction of two numbers,
a number and a expression or two expressions. In these cases, try to
simplify the ratio first. By the division property of radicals
^{3}√(x/y) = ^{3}√x / ^{3}√y.
Example:
Simplify the following.
^{3}√(8x^{3}/64y^{6})
By
the division property of radicals ^{3}√(8x^{3}/64y^{6})
= ^{3}√(8x^{3})
/ ^{3}√(64y^{6})
^{3}√(8x^{3})
= 2x
because (2x)^{3}
= 8x^{3}
^{3}√(64y^{6})
= 4y^{2}
because (4y^{2})^{3}
= 64y^{6}
Therefore
^{3}√(8x^{3}/64y^{6})
= (2x)/(4y^{2})
= x/(2y^{2}).
Example:
Simplify the following.
^{3}√(216y^{5}/8y^{8})
By
the division property of radicals ^{3}√(216y^{5}/8y^{8})
= ^{3}√(27/y^{3})
because 216/8 = 27 and (y^{5})/(y^{8})
= 1/(y^{3}).
^{3}√(27)
= 3 and ^{3}√(1/y^{3})
= 1/y.
Therefore
^{3}√(216y^{5}/8y^{8})
= 3/y.
• Note
that it's easier to solve cube roots if you know the first several
perfect cubes. The first ten perfect cubes are 1, 8, 27, 64, 125,
216, 343, 512, 729 and 1000.
We
reviewed how to take the square root and cube root of numbers and
expressions. There are also fourth roots, fifth roots, sixth roots,
seventh roots and so on. The procedure for solving these roots is the
same as for square roots and cube roots. For example, a fourth root
of a natural number x
is a number multiplied by itself four times to equal x,
and so on for higher roots. If n
is an odd natural number greater than 1 (n
> 1), then ^{n}√(x)
is an odd root. When
n
is an even natural number greater than 1 (n
> 1) and x
> 0, then ^{n}√(x)
is an even root.
Examples:
Simplify the following
radical expressions.
a.
^{4}√10000
^{4}√10000
= 10 since 10^{4}
= 10 ∙ 10 ∙ 10 ∙ 10 = 10000
b.
^{5}√(32x^{15})
^{5}√32
= 2 because 2^{5}
= 32
^{5}√(x^{15})
= x^{3}
because (x^{3})^{5}
= x^{15}
Therefore
^{5}√(32x^{15})
= 2x^{3}
c.
^{6}√(729x^{6}y^{18})
^{6}√729
= 3 because 3^{6} = 729
^{6}√(x^{6})
= x because (x)^{6} = x^{6}
^{6}√(y^{18})
= y^{3} because (y^{3})^{6} =
y^{18}
Therefore
^{6}√(729x^{6}y^{18})
= 3xy^{3}
d.
^{7}√(x + 10)^{14}
^{7}√(x
+ 10)^{14 }= (x + 10)^{2} because [(x +
10)^{2}]^{7} = (x + 10)^{14}
Review Problems: Set 1
Simplify
each of the following square roots.
1.
√121 2. √(4/25) 3. √0.49
4.
√60 5. √125 6. √252
Simplify
each of the following radical expressions.
7.
√(9y^{2}) 8. √(64x^{4})
9. √(x^{2} + 14x + 49)
10.
√(y^{2} – 8y + 16) 11. √(81x^{8}y^{6})
12. √(144x^{4}y^{2}z^{6})
13. √(80x^{5}) 14. √(75xy^{2}z^{3}) 15. √(98z^{3})
Simplify
each of the following cube roots.
16.
^{3}√64 17. ^{3}√(27/125)
18. ^{3}√216
19.
^{3}√40 20. ^{3}√320
21. ^{3}√54
Simplify
each of the following radical expressions.
22.
^{3}√(8x^{4}y^{5}z^{6})
23. ^{3}√(108z^{4}) 24. ^{4}√256
25.
^{5}√(243a^{5}) 26. ^{6}√(32x^{12})
27. ^{7}√(10^{7})
Rational Exponents
Just
as we can raise a number or expression to exponents that are
integers, we can raise them to fractional exponents as well. If n
is a positive integer greater then 1, then x^{1/n
}
is the nth
root of x
and
1/n
is a rational
exponent.
In other words x^{1/n}
= ^{n}√x.
To
make this more clear, consider √3 and 3^{1/2}.
By squaring both of these, we can determine that they are equal.
(√3)^{2}
= (√3)(√3) = √9 = 3
(3^{1/2})^{2}
= 3^{(1/2
∙ 2)}
= 3^{1}
= 3
Notice
that squaring both gives a result of 3. Therefore √3 = 3^{1/2}.
We
can rewrite exponential expressions in radical form and radical
expressions in exponential form. The radicand of a radical
expression becomes the base of an exponential expression and one over
the index of the radical expression becomes the exponent. We'll
illustrate this in the following
^{3}√6
= 6
^{1/}^{3}.
Notice the index in red becoming the denominator of the exponent and
the radicand in blue becomes the base.
Examples:
Evaluate
the following.
1.
16^{1/2
}=
√16 = 4. Recall that the index of a square root is 2, although the
2 is not written.
2.
(27)^{1/3
}=
^{3}√(27)
= 3 because (3)^{3
}=
27.
3.
(16/81)^{1/4}
= ^{4}√(16/81)
= 2/3 because (2/3)^{4}
= 16/81.
4.
32^{1/5
}=
2 because (2)^{5}
= 32.
Examples:
Simplify
the following rational expressions.
1.
(25x^{4})^{1/2
}
^{ }We
will break this down into the number part and the variable part.
25^{1/2
}=
5
and (x^{4})^{1/2
}=
√(x^{4})
= x^{2}
since (x^{2})^{2}
= x^{4}.
Therefore
(25x^{4})^{1/2
}=
5x^{2}.
• Note
that (x^{4})^{1/2
}can also be simplified using the rules for
raising an exponent to an exponent. Therefore, the exponents 4 and ½
are multiplied together to get 2, which is the new exponent and
(x^{4})^{1/2
}= x^{2}.
2.
(125x^{3})^{1/3
}
^{ }(125)^{1/3
}=
^{3}√(125)
= 5
since (5)^{3}
= 125 (x^{3})^{1/3
}=
^{3}√(x^{3})
= x
Therefore,
(125x^{3})^{1/3
}
= 5x.
3.
[(2a
+ 7)^{4}]^{1/4
}
= ^{4}√(2a
+ 7)^{4}
= 2a
+ 7
Notice
the answer is 2a
+ 7
and not 2a
+ 7. 2a
+ 7^{4}
= (2a
+ 7)^{4
}and
a
can be any number, therefore 2a
+ 7 could be a negative number. But we can't have a real number
solution if 2a
+ 7 is negative because no real number raised to the 4^{th}
power is negative. Therefore we must have absolute value to ensure a
positive real number solution.
4.
[(4b
– 3)^{5}]^{1/5
}=
^{5}√(4b
– 3)^{5}
= 4b
– 3.
In
this case, since n is
odd, there is no need for the absolute value symbols in the solution.
The value of 4b 3
could still be negative but a negative quantity raised to an odd
power will be negative.
In
the previous examples all of the exponents were in the form 1/n.
Now we will learn how to simplify exponential expressions where the
exponent is in the form p/n
where p and n
are positive integers and n ≠ 1.
To illustrate how to simplify , consider the following example.
27^{2/3}.
We can rewrite this in the form x^{1/}^{n}
raised to the 2nd power, (x^{1/}^{n})^{2}
= x^{2/}^{n}.
Therefore,
27^{2/3 }= (27^{1/3})^{2
}
^{
}=^{
} (^{3}√27)^{2}
= 3^{2} = 9
We
can rewrite the exponential expression in radical form. The
numerator of the rational exponent becomes the power of the radical
expression. The denominator of the rational exponent becomes the
index of the radical expression and the base of the exponential
expression becomes the radicand. In the previous example,
27^{2}^{/3}
= (^{3}√27)^{2}
We
would solve this as follows:
Take
the cube root of 27: ^{3}√27
= 3
Square
3: 3^{2} = 9
Or we can think of 27^{2/3 }as
27^{2}^{/3}
= ^{3}√(27)^{2}
We
would solve this as follows:
Square
27 first : 27^{2} =
729
Take
the cube root of 729 : ^{3}√729
= 9 since 9^{3} =
729.
• Note
that we know that 729 is divisible by 9 because the sum of the digits
in 729 is divisible by 9. The sum of the digits of 729 is 18, which
is divisible by 9.
Subscribe to:
Posts (Atom)