Page 81 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

√ √ √ √ √ √ √
On the other hand, multiply x + y = ab by x : x + xy = abx , thus abx
is a positive integer. Since a, b are distinct prime numbers, then x = abt , t ∈ N . Same
2
√ √ √
logic follows for y : y = abs , s ∈ N . Therefore, x + y = ab becomes
2
√ √ √ √ √
ab(t + s) = ab ⇒ t + s =1, a contradiction to t + s ≥ 2. As a result, x + y = ab
has no positive integer solution.

2.88 The real numbers x,y,z satisfy the equations

x + y + z =2,
xyz =4.

(1) Find the minimum value of the largest one of x,y,z ; (2) Find the minimum value of
|x| + |y| + |z| .

Solution: (1) Without loss of generality, assume x is the largest one among x,y,z , that is,

x ≥ y, x ≥ z . The first equation implies that x> 0 and y + z =2 − x , and the second
4
equation implies yz = , thus y, z are the two roots of the quadratic equation
x
2
u − (2 − x)u + 4 =0. The discriminant 2 4 3 2 2
x Δ=(2−x) −4· ≥ 0 ⇒ x −4x +4x−16 ≥ 0 ⇒ (x +4)(x−4) ≥ 0 ⇒ x−4 ≥
x
4
2
3
2
Δ=(2−x) −4· ≥ 0 ⇒ x −4x +4x−16 ≥ 0 ⇒ (x +4)(x−4) ≥ 0 ⇒ x−4 ≥ 0 ⇒ x ≥ 4.
2
x
0 ⇒ x ≥ 4
Hence, x =4 is the minimum value of the largest one of x,y,z . At this time, y = z = −1.
(2) Since xyz > 0, then x,y,z are all positive, or they are one positive two negative.
If x,y,z are all positive, (1) implies x ≥ 4, a contradiction to x + y + z =2.
If x,y,z are one positive two negative, without loss of generality we assume x> 0,y < 0,z < 0,
then |x| + |y| + |z| = x − y − z = x − (y + z) = x − (2 − x)=2x − 2. (1) implies x ≥ 4,
thus 2x − 2 ≥ 162x − 2 ≥ 16. x =4,y = z = −1 satisfy all conditions and the equal sign is obtained in the
inequality. Hence, the minimum value of |x| + |y| + |z| is 6.

2.89 a, b, c are nonzero real numbers, solve the system of equations

2
(x + y)(x + z)= a , (i)
2
(y + z)(x + y) = b , (ii)
2
(x + z)(y + z)= c . (iii)

Solution 1: (i)× (ii)/(iii), (i)× (iii)/(ii), (ii)× (iii)/(i) ⇒

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