Page 9 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

√
2b
−b 2
b
b
Solution: a + a −b =2 2 ⇒ (a + a ) =8 ⇒ a + a −2b =6 . Thus
b
−b 2
b
2b
(a − a ) = a − 2+ a −2b =6 − 2= 4 ⇒ a − a −b = ±2 .
b
b
The conditions a> 1,b > 0 imply that a − a −b > 0. As a conclusion, a − a −b =2.
1.11 Find the integer part of A = 11×70+12×69+13×68+···+20×61 × 100.
11×69+12×68+13×67+···+20×60
= 11×69+12×68+13×68+···+20×60 × 100
11×69+12×68+13×67+···+20×60
11+12+13+···+20
Solution: A = 11×69+12×68+13×68+···+20×60 × 100 + 11×69+12×68+13×67+···+20×60 × 100
11×69+12×68+13×68+···+20×60
× 100
=
11×69+12×68+13×67+···+20×60
11×69+12×68+13×67+···+20×60 = 100 + 11+12+13+···+20 × 100
11+12+13+···+20
+ + 11+12+13+···+20 × 100 × 100 11×69+12×68+13×67+···+20×60
11×69+12×68+13×67+···+20×60
11×69+12×68+13×67+···+20×60
11+12+13+···+20
11×69+12×68+13×67+···+20×60 .
= 100 + = 100 + 11+12+13+···+20 × 100 × 100
11×69+12×68+13×67+···+20×60
31 11 + 12 + 13 + ··· + 20
31 11 + 12 + 13 + ··· + 20 1 = × 100
Since 1 = × 100 69 (11 + 12 + 13 + ··· + 20) × 69
11 + 12 + 13 + ··· + 20
31 69 (11 + 12 + 13 + ··· + 20) × 69 11 + 12 + 13 + ··· + 20
1 = 11 + 12 + 13 + ··· + 20 × 100 < × 100
< 69 (11 + 12 + 13 + ··· + 20) × 69 × 100 11 × 69 + 12 × 68 + 13 × 67 + ··· + 20 × 60
11 × 69 + 12 × 68 + 13 × 67 + ··· + 20 × 60 11 + 12 + 13 + ··· + 20 2 .
11 + 12 + 13 + ··· + 20
11 + 12 + 13 + ··· + 20 2
< < × 100 = 1 × 100 < × 100 = 1
11 × 69 + 12 × 68 + 13 × 67 3 (11 + 12 + 13 + ··· + 20) × 60 3
(11 + 12 + 13 + ··· + 20) × 60 + ··· + 20 × 60
11 + 12 + 13 + ··· + 20 2
<
(11 + 12 + 13 + ··· + 20) × 60 A is 100 + 1 = 101.
Therefore the integer part of × 100 = 1 3
a + b
1.12 If a<b< 0 and a + b =4ab, evaluate .
2
2
a − b
√
2
2
Solution 1: a + b =4ab ⇒ (a + b) =6ab . Since a<b< 0, a + b = − 6ab. Similarly,
2
√
√
we can obtain a − b = − 2ab . Hence, a + b = − 6ab = √ 3 .
√
a − b − 2ab
x+y
Solution 2: Let a + b = x , a − b = y , then a + b = x ,b = x−y . Substitute them into
=
2 2
√ √
2
2
2
2
a + b =4ab to obtain x =3y . Since x, y < 0, then x = 3y, that is a + b = 3(a − b).
Thus a + b = √ 3 .
a − b
√ √
1.13 Given b, x =(b n+1 − a n+1 ) n+1 n x + n+1 a x 2 + n a + n+1 x a 2 − b
n
n
a,
n , compute the value of A =
n
n n
n
n n

√ √
n n+1 2 n n+1 2
n
n n
n n
A = x + a x + a + x a − b .
n

n n n+1 n n n n n 2 n n n n 2 n n n n n n 2
n
a,
Solution: b, x =(b n+1 − a n+1 ) n ⇒ x n+1 = b n+1 − a n+1 . Thus A = n n x 2 n+1 (x n+1 + a n+1 )+ n a 2 n+1 (a n+1 + x n+1 ) − b = n x n+1 + a n+1 n ( n x 2 n+1 +

n
n
n
n
n
n
n
)+
x n+1 (x n+1 + a n+1 a n+1 (a n+1 + x n+1 ) − b = x n+1 + a n+1 ( x n+1 +
A =
n n 2 n 2 n n n n n 2 n n n n n n n n n n n 2 n
n n
n
n x (x n x + a + a )+ n+1 (x a n+1 (a n+1 ) − b = ) − b b= n − a x n+1 + n+1 (b x − a + n+1 +
+ a a
+ a + x
n A =
n+1
) − b = n+1
n
n n 2 n n n n 2 n n n n n n 2 n a 2 n+1 n+1 n n+1 n+1 n n n+1 n+1 n+1 n n n+1 n n+1 n ( n+1 n+1 n
n n n
n ) − b =
+

A = n n x 2 n+1 (x n n+1 + a n n+1 )+ n n a 2 n+1 (a n n+1 + x n n+1 ) − b = n x n n+1 + a n n+1 n ( n x 2 n+1 a n+1 1x n+1 n + a n+1 (x n+1 + a n+1 ) − b = b n+1 − a n+1 + a n+1 (b n+1 − a n+1 +
A = x n+1 (x n+1 + a n+1 )+ a n+1 (a n+1 + x n+1 ) − b = x n+1 + a n+1 ( x n+1 + n n n a 2 n+1 ) − b = b n+1 n b n+1 − b = b − b =0 n n n n n

n
n
n
n
1
n
n
n+1 ) − b =
− b = b − b =0 a
n+1 n+1
n
n
n n 2 n n n n n n n n n a a n+1 ) − b = b x b n+1 + a n+1 (x n+1 + n+1 ) − b = b n+1 − a n+1 + a n+1 (b n+1 − a n+1 +

n n a 2 n+1 ) − b = n x n n+1 + a n n+1 (x n n+1 + a n n+1 ) − b = n b n n+1 − a n n+1 + a n n+1 (b n n+1 − a n n+1 + n 1 n
n ) − b =
a n+1 1x n+1 n + a n+1 (x n+1 + a n+1 ) − b = b n+1 − a n+1 + a n+1 (b n+1 − a n+1 +a n+1 ) − b = b n+1 b n+1 − b = b − b =0 .
a n n+1 ) − b = b 1 n+1 n b n+1 − b = b − b =0
a n+1 ) − b = b n+1 b n+1 − b = b − b =0
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