Page 55 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

k> 2 or k< −1
⇒ −2 <k < 4 ⇒ 3 <k < 4 or −2 <k < −1.
k> 3 or k< 0

√ √ √
2.19 Solve the equation x +2x − 63 + x +9 − 7 − x + x + 13 = 0.
2
2 2
Solution: The equation has real roots if and only if x +2x − 6 ≥ 0,x +9 ≥ 0, 7 − x ≥ 0 ⇒ x ≤ 7x +2x − 6 ≥ 0,x +9 ≥ 0, 7 − x ≥ 0 ⇒ x ≤ 7
3

2
x +2x − 6 ≥ 0,x +9 ≥ 0, 7 − x ≥ 0 ⇒ x ≥ ≤ 7 or x ≤−9,x ≥−9,x ≤ 7 ⇒ x = −9 or x =7. It is easy to obtain that x = −9 is a
root of the original equation, but x =7 is not. Hence, the original equation has a unique
root x = −9.

2
2.20 The equation k lg x + 3(k − 1) lg x +2k =0 has the variable x and the parameter
k , if the equation has two roots, one less than 100, one greater than 100, Find the range
of k .

Solution: Let t = lg x , and x 1 < 100,x 2 > 100, then the original equation becomes
kt + 3(k − 1)t +2k =0. Because x 1 < 100 <x 2 , we have
2

lg x 1 < 2 < lg x 2 ⇒ t 1 < 2 <t 2 ⇒ kf(2) < 0 ⇒ k[4k + 6(k − 1) + 2k] <
lg x 1 < 2 < lg x 2 ⇒ t 1 < 2 <t 2 ⇒ kf(2) < 0 ⇒ k[4k + 6(k − 1) + 2k] < 0 ⇒ 0 ⇒
k(2k − 1) < 0 ⇒ 0 <k < 1/2
k(2k − 1) < 0 ⇒ 0 <k < 1/2.

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