Page 84 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.91 Solve the equation x − [x]=3.
3
Solution: x =[x]+ {x}⇒ [x]= x −{x} , then the equation is equivalent to

3
3
3
x − (x −{x}) =3 ⇔ x − x =3 −{x} . Since 0 ≤{x} < 1, then 2 <x − x ≤ 3 ⇔ 2 < (x − 1)x(x + 1) ≤ 3
2 <x − x ≤ 3 ⇔ 2 < (x − 1)x(x + 1) ≤ 3 ( ). When x ≤−1, (x − 1)x(x + 1) < 0, ( ) has no solution. When
3
x ≥ 2, x − x = x(x − 1) ≥ 2(2 − 1) = 6, ( ) has no solution. When 1 <x< 2, [x] =1,
2
3
2
√
3
3
then the original equation becomes x − 1= 3 ⇒ x =4 ⇒ x = 3 4, which is the root of the
original equation.
x
x
2
2
2.92 The x -relevant equation (a − 1)( x−1 ) − (2a + 7)( x−1 +1 = 0 has real
roots. (1) Find the range of the parameter a . (2) If the equation has two real roots x 1 ,x 2,
3
and x 1 −1 + x 2−1 = 11 , find the value of a .
x 1
x 2
x
Solution: (1) Let x−1 = t , t =1, then the equation becomes (a − 1)t − (2a + 7)t +1 = 0.
2
2
When a − 1= 0, a = ±1, the equation is equivalent to −9t +1 = 0 or −5t +1 = 0, thus
2
1
1
t = or t = . When t = , x = whose root is x = − . When t = , x = whose
1
1
1
1
1
9 5 9 x−1 9 8 5 x−1 5
1
root is x = − . Hence, the original equation has real roots when a = ±1.
4
When a �= ±1, the equation (a − 1)t − (2a + 7)t +1 = 0 has real roots if and only if
2
2
Δ=(2a + 7) − 4(a − 1) = 28a + 53 ≥ 0 which implies a ≥− . When a = − , the
53
2
2
53
28 28
equation (a − 1)t − (2a + 7)t +1 = 0 has two identical roots which are not one. Hence,
2
2
when a ≥− , the original equation has real roots, that is, the range of a is [− , +∞).
53
53
28 28
(2) Since x 1 −1 , x 2 −1 are the two roots of (a − 1)t − (2a + 7)t +1 = 0, Vieta’s formulas
2
2
x 1
x 2
imply x 1 + x 2 = 2a+7 . On the other hand, we have x 1 + x 2 = 3 , thus
2
x 1 −1 x 2−1 a −1 x 1 −1 x 2−1 11
2a+7 = 3 3 ⇒ 3a − 22a − 80 = 0 ⇒ (a − 10)(3a + 8) = 0 ⇒ a = 10 8
2
− 80 = 0 ⇒ (a − 10)(3a + 8) = 0 ⇒ a = 10 or a = − . Since
2
2a+7
⇒ 3a − 22a
2
a −1 2 11 = 11 3
a −1
a ≥− , then a = 10 is the only possibility.
53
28
2.93 Solve the equation 2 log a + log a + 3 log 2 a =0.
ax
a x
x
Solution: If a =1, then the equation becomes 6 log 1 =0 whose solution set is x> 0 but
x
x =1.
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