Page 18 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

√ √
2
2
2
2
Solution: Let y = x− x − 1 ⇒ x−y = x − 1 ⇒ x −2xy+y = x−1 ⇒ x −(2y+1)x+y +1 =0 (i).
0
2
2
Δ=(2y + 1) − 4(y + 1) = 4y − 3 ≥ 0 ⇒ y ≥ 3/4 . Substitute the minimum value of
5
2
2
2
y , 3/4, into (i): x − x + 25 =0 ⇒ 1 (16x − 40x + 25) = 0 ⇒ (4x − 5) =0 ⇒ x =5/4.
2 16 16
√
Hence, when x =5/4, x − x − 1 has the minimum value 3/4.
1.45 If a, b, c are nonzero real numbers, and a + b + c = abc, a = bc , show a ≥ 3.
2
2
Proof: The conditions a + b + c = abc, a = bc imply that b + c = abc − a = a − a ⇒
2
3
3
b + c = a − a,
2
bc = a .
2
3
2
Hence, we can treat b, c as two roots of the quadratic equation x − (a − a)x + a =0.
Since a, b, c are nonzero real numbers, we have
Δ=(a − a) − 4a ≥ 0 ⇒ a (a + 1)(a − 3) ≥ 0 ⇒ a − 3 ≥ 0 ⇒ a ≥ 3.
2
2
2
2
3
2
2
2
1.46 t is a positive integer, show 2 and 3 are not common factors of t − t +1 and
2
t + t − 1.
2
Proof: t is a positive integer, thus one of the two consecutive integers t − 1 and t should
2
2
be an even number, then t − t = t(t − 1) is even, then t − t +1 is odd. Same logic to
2
2
2
get t + t is even, t + t − 1 is odd. Hence, 2 is not a common factor of t − t +1 and
2
t + t − 1.
One of the three consecutive integers t − 1,t,t +1 should be divisible by 3, thus at least
2
one of t − t = t(t − 1) and t + t = t(t + 1) is divisible by 3. Therefore, at least one of
2
t − t +1 and t + t − 1 is not divisible by 3.
2
2
1.47 If 3a − b +2c =8,a +4b − c =2, evaluate 6a + 11b − c .
Solution: Let 6a + 11b − c = m(3a − b +2c)+ n(a + 4b − c) = (3m + n)a + (4n − m)b + (2m − n)c ,

then equaling the coefficients to obtain

3m + n =6
4n − m = 11
2m − n = −1

⇒ m =1,n =3 ⇒ 6a + 11b − c =1 × 8+ 3 × 2 = 14.

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