Page 32 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1.84 The positive real numbers p, q satisfy p + q =7pq and make the polynomial
2
2
xy + px + qy +1 of x, y be factored into a product of two first-order polynomials, find
the values of p, q .

√
Solution: p> 0,q > 0,p + q =7pq ⇒ p +2pq + q =9pq ⇒ p + q =3 pq . Since the
2
2
2
2
polynomial xy + px + qy +1 can be factored into a product of two first-order
polynomials, we have xy + px + qy +1 = (ax + b)(cy + d)= acxy + adx + bcy + bd .
√
2
2
p> 0,q > 0,p + q =7pq ⇒ p +2pq + q =9pq ⇒ p + q =3 pq
Make the corresponding coefficients equal: ac =1, bd =1, ad = p, bc = q , thus 2 2
√ √ √ √ √
= abcd =1,p + q =3 pq =3 ⇒ p = 3+ 5 ,q = 3− 5 or p = 3− 5 ,q = 3+ 5 .
2 2 2 2
1.85 The positive integers a, b, c satisfy a + b + c +3 < ab +3b +2c , find the
2
2
2
values of a, b, c .
Solution: a, b, c are positive integers, thus a + b + c +3 and ab +3b +2c are both
2
2
2
integers, then a, b, c
2 + b + c +3 < ab +3b +2c ⇒ a + b + c +4 ≤ ab +3b +2c ⇒ a − ab + b 2 +
2
2
2
2
2
2
2 2 2 2 2 2 2 b 2
+ b + c +3 < ab +3b +2c ⇒ a + b + c +4 ≤ ab +3b +2c ⇒ a − ab + 4 + 2 2 4
2
2
b
2
2
2
2
2
3
2
2
2
b
b 2
+ b + c +3 < ab +3b +2c ⇒ a + b + c +4 ≤ ab +3b +2c ⇒ a − ab + 2
b
b 2
4
3b 2 −3b+3+ c −2c +1 ≤ 0 ⇒ (a− ) + (b−2) +(c −1) ≤ 0 ⇒ a− =0,b−2=+ + 3b 4 −3b+3+ c −2c +1 ≤ 0 ⇒ (a− ) + (b−2) +(c −1) ≤ 0 ⇒ a− =0,b−2=
2 2
2
2 2
3 2
2
b 2 2
2 + b + c +3 < ab +3b +2c ⇒ a + b + c +4 ≤ ab +3b +2c ⇒ a − ab +
2
2
4
2
3
b
b
4
2
4 2
2
2
4 2 3b −3b+3+ c −2c +1 ≤ 0 ⇒ (a− ) + (b−2) +(c −1) ≤ 0 ⇒ a− =0,b−2= 0,c − 1 =0 ⇒ a =1,b =2,c =1
2
2
3
b
2 2
4
3b
2
2
4
2
b
2
0,c − 1 =0 ⇒ a =1,b =2,c =1 (a− ) + (b−2) +(c −1) ≤ 0 ⇒ a− =0,b−2=
−3b+3+ c −2c +1 ≤ 0 ⇒
0,c − 1 =0 ⇒ a =1,b =2,c =1
4
4
2
2
0,c − 1 =0 ⇒ a =1,b =2,c =1.
1.86 The positive integers m, n satisfy (11111 + m)(11111 − n) = 123456789, show
m − n is a multiple of 4.
Proof: Since 123456789 is an odd number, then 11111 + m and 11111 − n are odd
numbers, then m, n are both even numbers.
(11111 + m)(11111 −n) = 123456789 ⇔ 11111 ×11111 −11111n+ 11111m−mn =
(11111 + m)(11111 −n) = 123456789 ⇔ 11111 ×11111 −11111n+ 11111m−mn =
123456789 ⇔ 11111(m − n) = mn + 2468
123456789 ⇔ 11111(m − n) = mn + 2468. Since mn is a multiple of 4 and
2468 = 4 × 617 is also a multiple of 4, then 11111(m − n) is a multiple of 4. In
addition, since 11111 and 4 are coprime, then m − n is a multiple of 4.
x y z
1.87 The nonzero real numbers a,b,c,x,y,z satisfy = = , evaluate
xyz(a + b)(b + c)(c + a) a b c
.
abc(x + y)(y + z)(z + x)
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