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Unformatted text preview: rvation leads to
x 1 x2 x 2 0
giving us the solutions x 1 or x 1 or x 2.
On the other hand, if we look for r and s to make
r3 s3 2
3rs 3
then we obtain
r3 3 1
r 2 which becomes
r 6 2r 3 1 0
which leads to the equation r 1. Thus r s 1 and we obtain the solutions
x r s 2
or
rs
3i r s
2 0 1.
x
2
2
d. x 3 6x 10 0
We look for numbers r and s such that
r 3 s 3 10
3rs 6
which gives us
r3 3 2
r 10 in other words,
10r 3 8 0 r6
and we obtain r3 5
and we may assume that r 3 5 17 and s x r s 3 17 17
3 17 . Thus we have the possibilities 5
5 3 17 5 or
x rs 3i r
2 s 3 5 17 3 5 17 3i 3 5 17 3 5 17 2 e. x 3 6x 2 6x 14 0. We begin by making the substitution u x
x 3 6x 2 6x 14 0 becomes 15 c and the equation . uc 6 uc 3 2 6 u c 14 0 giving us
u 3 3c 6 u 2 3c 2 12c 6 u c 3 6c 2 6c 14 0 and now we see that the definition c 2 will convert the equation into a form with no quadratic
term. We obtain
u 3 6u 10 0
We now copy the solution obtained in part d and obtain
u 17 3 5 17 5 3 or
3 u 5 17 3 5 17 3 5 17 3 3i 17 5 3 5 17 2 and so
x 2 17 3 5 or
3 x 2 5 17 3 5 17 3i 3 5 17 2 2. Show that if w 3 k 2 then the cubic equation
3wx 2k 0 x3
has one real solution
x 3 k2 w3 w3 k k2 3 k We look for numbers r and s such that
r 3 s 3 2k
3rs 3w
which gives us
r 3 k r 3 k k2 w3 s 3 k k2 w3 k2 w3 . In other words, we can assume that and and we conclude that the only real solution of the equation
x 3 3wx 2k 0
is
x 3 k2 w3 k 3 k2 w3 k . 3. By considering the solution of the equation
x 1 x2 x 3 0 (or otherwise) deduce that
3 9 3 5 11
63 16 93
3 5 11
63 1. 3 5 17
. We see at once that the only real solution of the equation
x 1 x2 x 3 0
is x 1. On the other hand, the equation can be written as
x 3 2x 3 0
and we can solve by looking for numbers r and s satisfying
r3 s3 3
3rs 2
From which we obtain
r3 3 2
3r 3 which gives us
r3 9 3 5 11
63 r 3 9 3 5 11
63 r 3 9 3 5 11
63 from which we can select and which gives us the real solution of the equation
x 1 x2 x 3 0
as
x r s 3 9 3 5 11
63 93
3 5 11
63 and we conclude that
3 4. 9 3 5 11
63 93
3 5 11
63 1. a. Show that if and are given numbers and if the equation
x x 2 x 0
is written in the form
x 3 ax b 0
then a
Since 2 and b .
x x 2 x x 3 2 x the desired result follows at once.
b. Given that and are real numbers and that 2 4, explain why the equation
x x 2 x 0
has only one real solution. By considering this equation, or otherwise, deduce that
3 3 2 2 4 2 3 63
With a 2 and b the inequality 17 3 3 2 2 3 63 4 2 . 27b 2 4a 3 0
says that
which holds if and only if
such that 2 2 2 2 2 4 4. Now suppose that 0 2 4. We look for two numbers r and s r 3 s 3
3rs 2 giving us
3 2
3r r3 27r 6 ...
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 Fall '08
 STAFF
 Math, Calculus

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