Multinomial Theorem 99
5.9 Multinomial Theorem
If n , n
1
, n
2
, . . . , n
k
are nonnegative integers and n = n
1
+ n
2
+ ···n
k
we put n
n
1
, n
2
··· n
k
= n
n
1
n
2
···n
k
. Using the De-Polignac Legendre Theorem, it is easy to see that this quantity is an integer. Proceeding in the same way we
proved the Binomial Theorem, we may establish the Multinomial Theorem: x
1
+ x
2
+ ··· + x
k n
=
X
n1+n2+···+nk=n
n
1
,n
2
,...,n
k
≥0
x
n
1
1
x
n
2
2
···x
n
k
k
.
We give a few examples on the use of the Multinomial Theorem.
660 Example Determine the coefficient of x
2
y
3
z
3
in x + 2y + z
8
. Solution: By the Multinomial Theorem
x + 2y + z
8
=
X
n1,n2,n3≥0
n
1
+n
2
+n
3
=8
8 n
1
, n
2
, n
3
x
n
1
2y
n
2
z
n
3
.
This requires n
1
= 2 , n
2
= 3 , n
3
= 3 . The coefficient sought is then 2
3
8 2
, 3, 3 .
661 Example In 1 + x
5
+ x
9 23
, find the coefficient of x
23
. Solution: By the Multinomial Theorem
X
n1,n2,n3≥0
n
1
+n
2
+n
3
=23
23 n
1
, n
2
, n
3
x
5n
2
+9n
3
.
Since 5n
2
+ 9n
3
= 23 and n
1
+ n
2
+ n
3
= 23 , we must have n
1
= 20 , n
2
= 1 , n
3
= 2 . The coefficient sought is thus
23 20
, 1, 2 .
662 Example How many different terms are there in the expansion of
x + y + z + w + s + t
20
? Solution: There as many terms as nonnegative integral solutions of
n
1
+ n
2
+ ··· + n
6
= 20 .
But we know that there are 25
5 of these.
Practice
100 Chapter 5
663 Problem
How many terms are in the expansion x + y + z
10
?
664 Problem Find the coefficient of x
4
in the expansion of 1 + 3x + 2x
3 10
?
665 Problem
Find the coefficient of x
2
y
3
z
5
in the expansion of x + y + z
10
?
Chapter
6
Equations
6.1 Equations in One Variable
Let us start with the following example.
666 Example Solve the equation 2
|x|
= sin x
2
. Solution: Clearly x = 0 is not a solution. Since 2
y
1 for y 0, the equation does not have a solution.
667 Example Solve the equation
|x − 3|
x
2
−8x+15 x−2
= 1 .
Solution: We want either the exponent to be zero, or the base to be 1. We cannot have, however, 0 as this is undefined. So,
|x − 3| = 1 implies x = 4 or x = 2. We discard x = 2 as the exponent is undefined at this value. For the exponent we want x
2
− 8x + 15 = 0 or x = 5 or x = 3 . We cannot have x = 3 since this would give 0
. So the only solutions are x = 4 and x = 5 .
668 Example
What would be the appropriate value of xif x
x
x. ..
= 2 made sense?
Solution: Since x
x
x. ..
= 2, we have x
2
= 2 the chain is infinite, so cutting it at one step does not change the value. Since we want a positive value we must have x =
√ 2.
669 Example
Solve 9 + x
−4
= 10x
−2
. Solution: Observe that
x
−4
− 10x
−2
+ 9 = x
−2
− 9x
−2
− 1 .
Then x = ±
1 3
and x = ±1.
670 Example Solve 9
x
− 3
x+1
− 4 = 0 .
Solution: Observe that 9
x
− 3
x+1
− 4 = 3
x
− 43
x
+ 1 . As no real number x satisfies 3
x
+ 1 = 0 , we discard this factor. So
3
x
− 4 = 0 yields x = log
3
4 .
101
102 Chapter 6
671 Example
Solve x − 5x − 7x + 6x + 4 = 504
. Solution: Reorder the factors and multiply in order to obtain
x − 5x − 7x + 6x + 4 = x − 5x + 4x − 7x + 6 = x
2
− x − 20x
2
− x − 42 .
Put y = x
2
− x . Then y − 20y − 42 = 504, which is to say, y
2
− 62y + 336 = y − 6y − 56 = 0 . Now, y = 6, 56, implies
x
2
− x = 6 and
x
2
− x = 56 .
Solving both quadratics, x = −2 , 4, −7, 8.
672 Example Solve 12x
4
− 56x
3
+ 89x
2
− 56x + 12 = 0 .
Solution: Reordering 12x
4
+ 12 − 56x
3
+ x + 89x
2
= 0 .
6.1 Dividing by x
2
, 12x
2
+ 1
x
2
− 56x + 1
x + 89 = 0
. Put u = x + 1
x. Then u
2
− 2 = x
2
+ 1 x
2
. Using this, 6 becomes 12u
2
− 2 − 56u + 89 = 0 , whence u = 52, 136. From
this x +
1 x
= 5
2 and
x + 1
x =
13 6
. Solving both quadratics we conclude that x = 1
2, 2, 23, 32.
673 Example Find the real solutions to
x
2
− 5x + 2
p
x
2
− 5x + 3 = 12 .
Solution: Observe that x
2
− 5x + 3 + 2
p
x
2
− 5x + 3 − 15 = 0 .
Let u = x
2
− 5x + 3 and so u + 2u
1 2
− 15 = u
1 2
+ 5u
1 2
− 3 = 0. This means that u = 9 we discard u
1 2
+ 5 = 0 , why?.
Therefore x
2
− 5x + 3 = 9 or x = −1 , 6.
674 Example Solve
p
3x
2
− 4x + 34 −
p
3x
2
− 4x − 11 = 9 .
6.2 Solution: Notice the trivial identity
3x
2
− 4x + 34 − 3x
2
− 4x − 11 = 45 .
6.3 Dividing each member of 8 by the corresponding members of 7, we obtain
p
3x
2
− 4x + 34 +
p
3x
2
− 4x − 11 = 5 .
6.4 Adding 7 and 9
p
3x
2
− 4x + 34 = 7 ,
from where x = − 5
3 , 3.
Practice 103
675 Example
Solve
3
√ 14 + x +
3
√ 14 − x = 4
. Solution: Letu =
3
√ 14 + x
, v =
3
√ 14 − x
. Then 64 = u + v
3
= u
3
+ v
3
+ 3uvu + v = 14 + x + 14 − x + 12196 − x
2 1
3
, whence
3 = 196 − x
2 1
3
, which upon solving yields x =
±13.
676 Example Find the exact value of cos 2
π
5. Solution: Using the identity
cosu ± v = cosucosv ∓ sinusinv
twice, we obtain cos 2
θ
= 2 cos
2
θ
− 1 6.5
and cos 3
θ
= 4 cos
3
θ
− 3 cos
θ
. 6.6
Let x = cos 2
π
5. As cos 6
π
5 = cos 4
π
5, thanks to 5 and 6, we see that x satisfies the equation 4x
3
− 2x
2
− 3x + 1 = 0 ,
which is to say x − 14x
2
+ 2x − 1 = 0 .
As x = cos 2
π
5 6= 1, and cos2
π
5 0, x positive root of the quadratic equation 4x
2
+ 2x − 1 = 0 , which is to say
cos 2
π
5 =
√ 5 − 1
4 .
677 Example How many real numbers x satisfy
sin x = x
100 ?
Solution: Plainly x = 0 is a solution. Also, if x 0 is a solution, so is −x 0. So, we can restrict ourselves to positive
solutions. If x is a solution then
|x| = 100|sinx| ≤ 100. So we can further restrict x to the interval ]0;100]. Decompose ]0;100] into 2
π
-long intervals the last interval is shorter: ]0; 100] =]0; 2
π
] ∪ ]2
π
; 4
π
] ∪ ]4
π
; 6
π
] ∪ ··· ∪ ]28
π
; 30
π
] ∪ ]30
π
; 100] .
From the graphs of y = sin x , y = x100 we that the interval ]0; 2
π
] contains only one solution. Each interval of the form ]2
π
k; 2k + 1
π
] , k = 1, 2, . . . , 14 contains two solutions. As 31
π
100, the interval ]30
π
; 100] contains a full wave, hence it contains two solutions. Consequently, there are 1 + 2
· 14 + 2 = 31 positive solutions, and hence, 31 negative solutions. Therefore, there is a total of 31 + 31 + 1 = 63 solutions.
Practice
104 Chapter 6
678 Problem
Solve for x 2
È
x a
+ 3
È
a x
= b
a +
6a b
.
679 Problem Solve
x − 7x − 3x + 5x + 1 = 1680 .
680 Problem Solve
x
4
+ x
3
− 4x
2
+ x + 1 = 0 .
681 Problem Solve the equation
2
sin2 x
+ 5 · 2
cos2 x
= 7 .
682 Problem If the equation
q
x +
È
x +
p
x + √
··· = 2 made sense, what would be the value of x?
683 Problem How many real solutions are there to
sin x = log
e
x?
684 Problem Solve the equation
|x + 1| − |x| + 3|x − 1| − 2|x − 2| = x + 2.
685 Problem Find the real roots of
p
x + 3 − 4 √
x − 1 +
p
x + 8 − 6 √
x − 1 = 1 .
686 Problem
Solve the equation 6x
4
− 25x
3
+ 12x
2
+ 25x + 6 = 0 .
687 Problem Solve the equation
x2x + 1x − 22x − 3 = 63 .
688 Problem
Find the value of √
30 · 31 · 32 · 33 + 1.
689 Problem Solve
x + √
x
2
− 1 x −
√ x
2
− 1 +
x − √
x
2
− 1 x +
√ x
2
− 1 = 98
.
690 Problem
Find a real solution to x
2
− 9x − 1
10
+ 99x
10
= 10x
9
x
2
− 1 .
Hint: Write this equation as x
2
− 9x − 1
10
− 10x
9
x
2
− 9x − 1 + 9x
10
= 0 .
691 Problem
Find the real solutions to
É
x + 2
q
x + 2
È
x + ··· + 2
p
x + 2 √
3x
| {z
}
n radicals
= x .
692 Problem Solve the equation
1 1 +
1 1 +
1 1 +
. .
. 1 +
1 x
= x .
where the fraction is repeated n times.
693 Problem Solve for x
p
x + √
x + 11 +
p
x + √
x − 11 = 4 .
6.2 Systems of Equations
Kategori