Page 89 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2
Solution: Let x = u, y = v, z 1/2 = t , then the system becomes

2
2
u + v + t 2 = 18 (i)
uv − vt = −3 (ii)
ut = 4 (iii)

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2
2
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(i)+(ii)×2-(iii)×2: u +v +t +2(uv−vt−ut) =4 ⇔ (u+v−t) =4 ⇒ u+v−t = ±2 ⇒ u−t = −v±2
u +v +t +2(uv−vt−ut) =4 ⇔ (u+v−t) =4 ⇒ u+v−t = ±2 ⇒ u−t = −v±2, substitute it into (ii): v ± 2v − 3= 0 whose roots are v = ±1 or v = ±3.
2
2
2
2
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Substitute the values of v into (ii)(iii):
u − t = −3
ut =4
t − u = −3
ut =4

3u − 3t = −3
ut =4
3t − 3u = −3
ut =4

Solve them to obtain (u,v,t)=(1, 1, 4), (−4, 1, −1), (4, −1, 1), (−1, −1, −4),
(u,v,t)=(1, 1, 4), (−4, 1, −1), (4, −1, 1), (−1, −1, −4), √ √ √
(u,v,t)=(1, 1, 4), (−4, 1, −1), (4, −1, 1), (−1, −1, −4),
√
√
√
√
√
√ √ √ √ ( √ √ 17−1 , 3, √ √ 17+1 ), ( √ √ 17+1 , 3, 17−1 ), ( 17+1 , −3, 17−1 ), ( 17−1 , −3, 17+1 )
√ √
√ √
√ √
17−1
17+1
17+1
17−1
17+1
17+1
17−1
17−1
2
2
, −3,
, −3,
), ( (
2
( ( 17−1 , 3, 3, 17+1 ), 17+1 , , 3, 3, 17−1 ), 17+1 , −3, 2 17−1 ), 2 17−1 , −3, 2 17+1 ) ) . Notice that the
2
), ( (
,
2
), ( (
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
second, fourth, sixth, eighth solutions have negative u = x which is impossible,
2
therefore all possible solutions of the original system are
2.100 Find the polynomial p(x) defined on a set of real numbers such that
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2
p(0) = 0 and p(x + 1) = [p(x)] +1.
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2
2
Solution: Let x =0 and substitute into p(x + 1) = [p(x)] +1 to obtain p(1) = [p(0)] +1 = 1
since p(0) = 0. Choose x =1, 2 to obtain p(2) = [p(1)] +1 = 2,p(5) = [p(2)] +1 = 5.
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2
2
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Keep going, we have p(26) = [p(5)] + 1 = 26,p(26 + 1) = [p(26)] + 1 = 26 +1, ··· .
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Hence, the equation p(x) − x =0 has infinitely many roots: 0, 1, 2, 5, 26, 26 +1, ···. Since
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p(x) − x is a polynomial, p(x) − x =0 always holds, that is, p(x) = x .
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