Page 17 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Hence, xy has the maximum value 6 and the minimum value −8.

1.41 If the real numbers a, b, c satisfy a + b + c =0, abc =1, show one of a, b, c

should be greater than 3/2.

Proof: Since a, b, c are real numbers and abc =1, we have at least one of a, b, c is greater
than zero. Without loss of generality, let c> 0.

a+b 2 a−b 2
1
a + b + c =0 ⇒ a + b = −c (i); abc =1 ⇒ ab = (ii); ab = − (iii).
c 2 2
3
1 c 2 a − b 2 a − b 2 c 2 1 c − 4
Substitute (i),(ii) into (iii): = − ⇒ = − = ≥ 0, which
c 4 2 2 4 c 4c
√
3 3 3 3
together with c> 0 implies c ≥ 4. Hence, c ≥ 4= 32/8 > 27/8=3/2.
1.42 Given m + n + p = 30 , 3m + n − p = 50 , m, n, p are positive, and x =5m +4n +2p,

find the range of x .

Solution: (3m + n + p) − (m + n + p) = 50 − 30 ⇒ m − p = 10 ⇒ m = 10 + p> 10 since
p> 0.

n n
(3m + n + p) + (m + n + p) = 50 + 30 ⇒ m + = 20 ⇒ m = 20 − < 20 since n> 0.
2 2
n + p = 30 − m ⇒ 10 <n + p< 20.

Hence, x =5m+4n+2p = (4m+2n)+(m+n+p)+n+p = 80+30+n+p = 110+n+p ∈
x =5m+4n+2p = (4m+2n)+(m+n+p)+n+p = 80+30+n+p = 110+n+p ∈
(120, 130)
(120, 130).
√
2
1.43 Given a, b, c are real numbers and satisfy a + b =4, 2c − ab =4 3c − 10, find
the values of a, b, c .

2 2
√ √ a − b
2
2
a + b
2
2
2
a + b
Solution: 2c − ab =4 3c − 10 ⇒ 2c − 4 3c + 10 = ab = a − b 2 − =
a − b
a + b
√ √
√ √
2
2
2c − ab =4 3c − 10 ⇒ 2c − 4 3c + 10 = ab = − − 2 = = 2
2
2
2c − ab =4 3c − 10 ⇒ 2c − 4 3c + 10 = ab =
2
2 2 2 2 2 2 2
√
4 4 − a − b − b =4 − a − b ⇒ 2c −4 3c +6+ a − b =0 ⇒ 2(c − 3) + 2 2
2
2 2 4 2 2 a − b 2 2 a − b √ √ 2 2 a − b √
a − b
=0 ⇒ 2(c − 3) +
⇒ 2c −4 3c +6+
a − b
√
2 √
2
=4 −

a −
2
2 −
2
2 − 2 2 2 2 2 =4 2 2 ⇒ 2 2c −4 3c +6+ 2 2 =0 2 ⇒ 2(c − 3) +

a − b 2
a − b √ √
a − b =0 ⇒ c = √ 3,a = b =2
3,a =
=0 ⇒ c = b =2
2
2 =0 ⇒ c = 3,a = b =2.
2
√
1.44 x is a real number, find the minimum value of x − x − 1 and its associated x value.
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