Page 24 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Proof: x + y + z = a ⇒ z = a − (x + y) and substitute it into x + y + z = a /2 to obtain
2
2
2
2
2 2 2 2 2 2 2 2
x + y + (x + y) − 2a(x + y)+ a = a /2 ⇒ x + y + xy − ax − ay + a /4= 0 ⇒
2
2
2
2
2
2
2
2
x + y + (x + y) − 2a(x + y)+ a = a /2 ⇒ x + y + xy − ax − ay + a /4= 0 ⇒
2
2
y +(x − a)y +(x − a/2) =0
2
2
y +(x − a)y +(x − a/2) =0. Since x, y are real numbers, then Δ=(x − a) − 4(x − a/2) ≥ 0 ⇒ x(2a − 3x) ≥ 0 ⇒ 0 ≤ x ≤ 2a/3
2
2
Δ=(x − a) − 4(x − a/2) ≥ 0 ⇒ x(2a − 3x) ≥ 0 ⇒ 0 ≤ x ≤ 2a/3 . Similarly, we can show 0 ≤ y ≤ 2a/3, 0 ≤ z ≤ 2a/3.
2
2

Therefore x,y,z are nonnegative and not greater than 2a/3.
1.64 Two real numbers x, y satisfy x + y =2. Find the maximum value of x + y .
3
3
.
3
a,
2
3
3
2
Solution: Let x + y = t. b, x + y =2 ⇒ (x + y)[(x + y) − 3xy]= 2 ⇒ t(t − 3xy)=2 ⇒ xy = t −2
3t
3
t −2
2
3
=0
2
Thus we can treat x, y as the two roots of the quadratic equation tu + − tu + t −2 =0, then
−
3t 3t
3
3
4t − 8 −t +8
2
Δ= t − ≥ 0 ⇒ ≥ 0 ⇒ 0 <t ≤ 2 ⇒ 0 <x + y ≤ 2. Hence, the maximum value
3t 3t
of x + y is 2.
x +4
1.65 Write as partial fractions.
3
x +2x − 3
Solution: x =1 is a root of the cubic equation x +2x − 3 =0, thus x − 1 is a factor of
3
x +2x − 3 . Use polynomial long division to obtain the other factor x + x +3. Let
3
2

2
x +4 A Bx + c (A + B)x +(A − B +2C)x +3A − C . Make the coefficients
= + =
3
3
x +2x − 3 x − 1 x + x +3 x +2x − 3
2
equal to obtain
A + B =0
A − B + C =1
3A − C =4
x +4 1 x +1
⇒ A =1,B = −1,C = −1. Hence, = − .
3
2
x +2x − 3 x − 1 x + x +3
1.66 Show a + a + a − 1 is an integer for any positive integer a , and it has a
1
3
3 2
2 2
remainder of 2 when divided by 3.
2
3
3
1
3 2
Proof: a + a + a − 1= 2a +3a +a − 1= a(a+1)(2a+1) − 1. For any positive integer a ,
2 2 2 2
a(a + 1)
is an integer, thus a + a + a − 1 is an integer.
1
3
3 2
2 2 2
3
3 2
1
a + a + a − 1= a(a+1)(2a+1) − 1= 2a(2a+1)(2a+2) − 1. One of 2a, 2a +1, 2a +2 is a
2 2 2 8
multiple of 3. Since 3 and 8 are coprime, then 2a(2a + 1)(2a + 2) is divisible by 3. Hence
8
the original expression is a multiple of 3 minus 1, i.e. it has a remainder of 2 when divided
by 3.
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