Page 57 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Solution: Let the equilateral triangle has the side length a , then the radius of the inscribed
√ √
circle and the radius of the circumcircle are 3 a and 3 a , respectively. Vieta’s formulas
6 3
√ √ √ √ √ √ √ √
q
4
p
imply a + 3 a + 3 a = − (i), a 3 a + a 3 a + 3 a · 3 a = (ii), a · 3 a · 3 a = (iii).
6 3 3 6 3 6 3 3 6 3 3
√ √
(iii) leads to a =2, substitute it into (i)(ii) to obtain p = −6 − 3 3,q = 2 +6 3.
2.25 Solve the equation system
log x + log 8=2,
2
y
log 2 + log x 2 =1.
y
8
Solution: The system is equivalent to
3
log x + = 2 (i)
2
log y
2
1 2 log x
+ 2 = 1 (ii)
log y 3
2
(ii)×3-(i): log x =1 ⇒ x =2. Substitute it into (i): y =8. We can easily verify x =2,y =8
2
is a solution of the original system.

2.26 Given f (x) = x − , solve the equation f[f(x)] = x .
1
x
4
2
1 1 x −3x +1
Solution: f [f(x)] = x − − 1 = 3 ,
x x− x −x
x
√
2
4
x −3x +1 2 1 2
thus f [f(x)] = x ⇔ x −x = x ⇒ x = 2 ⇒ x = ± 2 .
3
2.27 n is a positive integer, and denote a n as the number of nonnegative integer
solutions (x,y,z) to the equation x + y +2z = n . Find the values of a 3 and a 2001 .
Solution: When n =3, we have x + y +2z =3. Since x ≥ 0,y ≥ 0,z ≥ 0, we have
0 ≤ z ≤ 1. When z =1, then x + y =1, then (x, y) = (0, 1) or (1, 0). When z =0,
then x + y =3, then there are four possibilities of (x, y). Hence, a 3 =2 + 4 =6.

When n = 2001, we have x + y +2z = 2001, thus 0 ≤ z ≤ 1000. When z =0, then
x + y = 2001, then there are 2002 possibilities of (x, y). When z =1, then

x + y = 1999, then there are 2000 possibilities of (x, y). ... .. . When z = 1000, then
x + y =1, then there are two possibilities of (x, y). As a conclusion,

a 2001 = 2002 + 2000 + 1998 + ··· + 4 + 2 = 1003002.

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