Page 78 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2
Proof: Let f(x) = ax + bx + c , then
3 3
3 3
9 9
(9a + 12b + 16c)(a + b +
f( ) · f(1) = ( a + b + c)(a + b + c) = =
f( ) · f(1) = ( a + b + c)(a + b + c) 1 1 (9a + 12b + 16c)(a + b + c) c). Since
16
16
4 4 16 4 4 16
√ √ √ √ √ √ √ √ √

2a + 3b + 5c =0, a, b= − 6a− 15c , then (9a + 12b + 16c)(a + b + c)=(9a − 4 6a − 4 15c + 16c)(a − √ 6 a − √ 15 c + c)=
√
√
3 √(9a + 12b + 16c)(a + b + c)=(9a − 4 6a − 4 15c + 16c)(a − √ a − 3 15 √ c + c)=
√ √
√
36 √
√
√
√
√
√
√
3 6
a
2
√ (9a +
√
√
√ √ √ √ [( 81 − √12b + 16c)(a + b + 240)c][ 3− 6 a + 3− 15 c]= c [( 81 − √ a − 3 15 √ c + c)=
√
√
96)a +( 256 − √c)=(9a − 4 6a − 4 15c + 16c)(a −
96) +( 256 −
3
3 15
2
240)c][
√
√
√
√
(9a + 12b + 16c)(a + b + c)=(9a − 4 6a − 4 15c + 16c)(a − √ 6 a − √ 15 c + c)= √ [( 81 − √ √ √ √ √ 3− 3 6 √ a + 3− √ c]= c [( 81 − √ c a 3 √
96)a +( 256 −
96) +( 256 −
c a
3− 15
3 6
2
3− 15
√ 96)a
] < 0
√
√
√
√
√
240)][−
√
√
√
√ (9a + 12b + 16c)(a + b + c)=(9a − 4 6a − 4 15c + 16c)(a − √ a − 315 √ c + c)=[( 81 3− 6 a + +( 256 − 240)c][ 3 3− 6 a + 3 3 c]= c [( 81 − 96) +( 256 −
c
3
a
3
2
3 15
3 6 a
3
√
√
√
√
√
√
[( 81 − √ 96)a +( 256 − √ 240)c][ 3− 6 a + 3− 15 c]= c [( 81 − √ 96) +( 256 − 240)][ 3− √ c + 3− √ ] < 0 , thus
√
3− 15
3 3− 6 a
3 6
ca
2
√
√ [( 81 − √ 96)a +( 256 − 240)c][ 3− a + 3− 3 15 c]= c [( 81 − 96) +( 256 − 240)][ 3 c c + 3 3 ] < 0
√
√
√ 240)][ 3− 6 a + 3− 15 ] < 0 3 3 c
3 6 a
240)][ 3− c + 3− 3 15 ] < 0 3 3
3 c 3 f( ) · f(1) < 0 , which implies that one root is within ( , 1).
4 4
2.84 The real numbers a, b satisfy
ax + by =3,
2
ax + by 2 =7,
3
ax + by 3 = 16,
4
ax + by 4 = 42,
5
5
compute ax + ay and x, y .
Solution 1: We have
(ax + by)(x + y)= ax + axy + bxy + by =(ax + by ) + (a + b)xy ;
2
2
2
2
(ax + by )(x + y)= ax + ax y + bxy + by =(ax + by ) + (ax + by)xy ;
2
2
3
2
2
3
3
3
(ax + by )(x + y)= ax + ax y + bxy + by =(ax + by ) + (ax + by )xy ;
4
2
2
4
3
3
4
3
4
3
(ax + by )(x + y)= ax + ax y + bxy + by =(ax + by ) + (ax + by )xy .
4
5
5
4
5
4
3
5
4
3
Substitute the given equations into them:
3(x + y) = 7 +(a + b)xy (i),
7(x + y) = 16 + 3xy (ii),
16(x + y) = 42 + 7xy (iii),
5
5
42(x + y) = (ax + by ) + 16xy (iv).
(ii)×7− (iii)×3: x + y = −14, substitute it into (ii): xy = −38. Substitute
x + y = −14, xy = −38 into (iv): ax + by = 42(−14) − 16(−38) = 20.
5
5
√ √
In addition, x + y = −14, xy = −38 ⇒ x = −7 − 87,y = −7+ 87 or
√ √
x = −7+ 87,y = −7 − 87.
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