Page 63 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Solution: The equation ||x − a|− b| =5 is equivalent to |x − a|− b = ±5 ⇔|x − a| = b ± 5.
The equation has three distinct roots if and only if b − 5= 0, that is b =5 and the roots are

x = a, x = a ± 10.

2.42 Solve the system of equations
2
x + xy + y 2 = 84,
√
x + xy + y = 14.
√ 2 2 √
Solution: The second equation is equivalent to x + y = 14 − xy ⇒ x + xy + y = 196 − 28 xy .
√ √
The left hand side is 84 due to the first equation, then 84 = 196 − 28 xy ⇒ xy =4.
Substitute it into the second equation to obtain x + y = 10. Hence, we can treat x, y as
roots of the quadratic equation z − 10z + 16 = 0. The roots are z =2 or z =8. Therefore,
2
the original system of equations has two solutions (2, 8), (8, 2).

√
2.43 The real numbers a, b, c satisfy a = b and 2009(a − b)+ 2009(b − c) + (c − a) =0 ,

find the value of (c − b)(c − a) .
(a − b) 2
√
2
Solution: Let 2009 = x , then (a − b)x +(b − c)x +(c − a) =0. a = b implies that
√
this equation is a quadratic equation. Obviously, x = 2009 and 1 are two roots of this

√ c − b √ c − a
quadratic equation, then 2009 + 1 = , 2009 × 1= .
a − b a − b
(c − b)(c − a) c − b c − a √ √ √
Hence, = × =( 2009 + 1) 2009 = 2009 + 2009.
(a − b) 2 a − b a − b

2.44 Find all functions f(x) that satisfy the equation 2f(1 − x) +1 = xf(x).

Solution: Replace x with 1 − x in the equation: 2f(x) + 1 = (1 − x)f(1 − x) (i).

Rewrite the original equation as f(1 − x) = [xf(x) − 1] (ii). Substitute (ii) into (i):
1
2
1 2
2 2
1 1
2f(x) + 1 = (12f(x) + 1 = (1
2f(x) + 1 = (1 − x) [xf(x) − 1] ⇔ 4f(x) +2 = xf(x) − 1 − x f(x)+ x ⇔
2 − x) [xf(x) − 1] ⇔ 4f(x) +2 = xf(x) − 1 − x f(x)+ x ⇔− x) [xf(x) − 1] ⇔ 4f(x) +2 = xf(x) − 1 − x f(x)+ x ⇔
2 2
x−3
2
(x − x + 4)f(x) = x − 3 ⇔ (x) = x −x+4 x) = [xf(x) − 1]
2 2
x−3 1
x−3
f
x −x+4 2 .
(x − x + 4)f(x) = x − 3 ⇔ (x) (1 (x)
2 = −=
(x − x + 4)f(x) = x − 3 ⇔
2
2
x −x+4
2.45 If the equality ab = 2(c + d) is always valid, show at least one of the equations
x + ax + c =0 and x + bx + d =0 has real root(s).
2
2
2
Proof: Δ 1 = a − 4c, Δ 2 = b − 4d . Assume Δ 1 < 0, then a − 4c< 0 , then a < 4c .
2
2
2
2
2
ab = 2(c + d) ⇔ ab − 2c =2d , thus Δ 2 = b − 4d = b − 2ab +4c> b − 2ab + a =(b − a) ≥ 0.
2
2
2
Similarly if we assume Δ 2 < 0, then we will obtain Δ 1 ≥ 0.
2.46 Solve the equation log x + log b =1 where a> 1,b > 1.
x
a
Download free eBooks at bookboon.com
63