Unformatted text preview: 27r 3 2 3 0 and we can take
4 r 3 s 3 2 4 3 3 2 2 2 6 3 and
3 3 2 2
6 3 then the real solution of the equation
x 2 x 0 x
appears as
r s 3 3 2 2 4 2 3 63 3 3 2 2 4 2 3 63 . Since that real solution has to be we obtain
3 3 2 2 4 2 3 63 3 3 2 2 4 3 63 2 . c. Given that and are real numbers and that 2 4, consider the two complex roots of the equation
x x 2 x 0
and deduce that
3 3 2 2 4 2 3 63 1
2
2 4
2 2 4 3 and
3 3 2 2 4 2 3 63
The complex roots are
rs 3i r
2 s and since these two complex roots are also
18 i 4
2 2 1
2 2
3 we deduce that
rs
and
4 s r By looking at the values of r and s we see that r
actually 3
s 0 and so the preceding equation is
4 s r 2 2
3 . Adding we obtain
4 2r 2
3 which gives us
3 3 2 2 4 2 3 63 1
2 4 1
2
2 4
2 2
3 and by subtracting we obtain
3 3 2 2 4 2 3 63 2
3 Some Exercises on the Trigonometric Method
1. By considering the equation
x 2 x3 0 1x show that
x3
27
cos
3 7x 6 0 1 arccos
3 93
77 2 and
27
cos
3 1 arccos
3 93
77 2
3 3 27
cos
3 1 arccos
3 93
77 2
3 1 and Solution: We observe first that
x 1x 2 x 3 x3 7x 6. To solve the equation
x 3 7x 6 0
we look for a solution of the form x k cos and express the equation as
k 3 cos 3 7k cos 6
and we choose k in such a way that 19 k3 4
3
7k
which will hold when
27
.
3 k
The equation 7x 6 0 x3
now becomes
3 27
3 cos 3 27
3 7 cos 6 which gives us
4 cos 3 93
77 3 cos in other words,
93
77 cos 3
and we deduce that either
x 27
cos
3 93
77 1 arccos
3 or
x 27
cos
3 1 arccos
3 93
77 2
3 x 27
cos
3 1 arccos
3 93
77 2
3 or Finally we observe that since
0 1 arccos
3 1 arccos
3
3 93
77
93
77 we have
cos 1
2 and
cos 93
77 1 arccos
3 2
3 0 2
3 1
2 and
cos 93
77 1 arccos
3 Thus the largest of the solutions of the equation is
27
cos
3 1 arccos
3 93
77 and the negative solution is
27
cos
3 1 arccos
3 20 93
77 2
3 and the remaining solution is
27
cos
3 1 arccos
3 93
77 2
3 2. Given real numbers , and , prove that the number
2 2 2
3 1 arccos
3 cos is the largest of the three numbers and and 3 3 2 2
2 2 2 2 2 . Hint: Apply the method of Exercise 1 to the equation
x x x 0. Exercises on Proof by Contradiction
1. Prove that the following numbers are irrational:
a. log 10 5
To obtain a contradiction, assume that log 10 5 is rational and choose positive integers m and n
such that
log 10 5 m
n
Since
10 m/n 5
we have
10 m 5 n
which is impossible because 10 m is even and 5 n is odd.
b. log 12 24 Solution: To obtain a contradiction, assume that the number log 12 24 is rational and choose
positive integers m and n such that
log 12 24 m .
n
Since
12 m/n 24
we have
12 m 24 n
which we can express as
12 m n 2 n
Since 2 n 1 we see that m n and so the...
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 Fall '08
 STAFF
 Math, Calculus

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