Page 23 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

b + c b + b
Solution 1: When a = b , we have (b − c) =0 ⇒ b = c , then = =2.
2
a b
When a = b , the given equality becomes
2 2 2 2
2
2
2
2
(b − c) = 4(a − b)(c − a) ⇒ (b + c) − 4a(b + c) +4a =0 ⇒ [(b + c) − 2a] =0 ⇒
(b − c) = 4(a − b)(c − a) ⇒ (b + c) − 4a(b + c) +4a =0 ⇒ [(b + c) − 2a] =0 ⇒
b + c b + c
b + c =2a ⇒ =2 =2 .
a
b + c =2a ⇒ a
Solution 2: When a = b, it is the same as solution 1.
When a = b , (b − c) − 4(a − b)(c − a)=0. Treat this as the discriminant of the quadratic
2
equation (a − b)x +(b − c)x +(c − a) =0. Since the sum of all coefficients is 0, then 1
2
is a root of this quadratic equation. Since Δ=(b − c) − 4(a − b)(c − a) =0 , then
2
(
)
)
)(
(
c − a b + c
x 1 = x 2 =1. Vieta’s formulas implies that x 1 x 2 = =1 ⇒ b+c =2a ⇒ = 2.
a − b a
1.62 Find the minimum positive integer A and the corresponding positive integer
B such that (1) A is divisible by 200 and its quotient divided by 19 has a remainder of

2, divided by 23 has a remainder of 10; (2) B >A , B − A has four digits and is divisible

by 3,4,17,25.

Solution: (1) ⇒ A/200 = 19U + 2 = 23V + 10 where U, V are positive integers, then

U, V = V + 4V +8 . Since U is a positive integer, then 4V +8 is a positive integer. Let
19
19
3
4V +8 = p , then V =4p − 2+ p in which p should be a positive integer. Since 3 and
3
U,
19 4 4
4 are coprime, then p =4n (n is a positive integer). To have minimum A, we choose
n =1,p =4,V = 17, then A = 200(23 × 17 + 10) = 80200.

According to (2) and since 3,4,17,25 are coprime, then B − A should be

3 × 4 × 17 × 25 × k = 5100k (k is a positive integer). In addition, B − A is a four-digit
number, thus k =1. Hence, B = A + 5100 = 85300.

1.63 If the real numbers x,y,z satisfy x + y + z = a , x + y + z = a /2 (a> 0),
2
2
2
2
show x,y,z are nonnegative and not greater than 2a/3.

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