Page 54 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.15 Solve the equation x − 4x +1 = 0.
4
2
Solution: Let y = x , then the equation becomes
2
√ √ √
2 2 2 2
y − 4y +1 = 0 ⇔ (y − 2) =3 ⇒ y =2 ± 3 ⇒ x = 2+ 3 or x =2 − 3,
√ √
the first of which implies x = ± 2+ 3= ± √ 3+1 , the second of which implies
2 3 2
√ √ √ √ √ √
x = ± 2 − 3= ± √ 3−1 . Hence, the four solutions are √ 3+1 , − √ 3+1 , √ 3−1 , − √ 3−1 .
2 3 2 2 3 2 2 3 2 2 3 2 2 3 2
2.16 For any real number k, the equation (k + k + 1)x − 2(a + k) x + k +3ak + b =0
2
2
2
2
always has the root x =1. Find (1) the real numbers a, b ; (2) the range of the other root
when k is a random real number.
Solution: (1) x =1 is always a root, then (k + k + 1) − 2(a + k) + k +3ak + b =0 is
2
2
2
always valid for any k , that is, (1 − a)k + (1 − 2a + b) =0 for any k , thus
1 − a =0, 1 − 2a + b =0, which lead to a = b =1.

(2) Let the other root be x 2 , then Vieta’s formulas imply

2
2
k +3ak + b k +3k +1
2
1 · x 2 = = ⇔ (x 2 − 1)k +(x 2 − 3)k +(x 2 − 1) = 0.
2
2
k + k +1 k + k +1
Δ=(x 2 − 3) − 4(x 2 − 1) = −3x +2x 2 +5 ≥ 0 which implies −1 ≤ x 2 ≤ 5/3.
2
2
2
2
2.17 Solve |3x −|1 − 2x|| =2.
Solution: |3x −|1 − 2x|| =2 ⇒ 3x −|1 − 2x| = ±2.
When 3x −|1 − 2x| =2, |1 − 2x| =3x − 2, then 3x − 2 ≥ 0 ⇒ x ≥ 2/3, and
1 − 2x = ±(3x − 2) which leads to x =1 or x =3/5 < 2/3 (deleted).

When 3x −|1 − 2x| = −2, |1 − 2x| =3x +2, then 3x +2 ≥ 0 ⇒ x ≥−2/3, and
1 − 2x = ±(3x + 2) which leads to x = −1/5 or x = −3 < −2/3 (deleted).

Hence, the original equation has solutions x =1 or x = −1/5.

2.18 The equation 7x − (k + 13)x + k − k − 2 =0 (k is a real number) has two real
2
2
roots α, β , and 0 <α < 1, 1 <β < 2. Fine the range of k .

2
Solution: Let f(x)=7x − (k + 13)x + k − k − 2, since 0 <α < 1, 1 <β < 2 are two
2
roots of f(x)=0, then
2
f(0) = k − k − 2 > 0
2
f(1) = k − 2k − 8 < 0
2
f(2) = k − 3k> 0

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