Page 62 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.37 If the equation x − kx + k − 4= 0 has two positive roots, find the range of k .
2
2
Solution: The condition of two positive roots (denoted by x 1 ,x 2)
implies x 1 + x 2 = k> 0,x 1 x 2 = k − 4 > 0, Δ= k − 4(k − 4) = −3k + 16 ≥ 0.
2
2
2
2
√
From these three inequalities, we can easily obtain 2 <k ≤ 4 3/3.
2.38 Solve the system of equations
√
2
2 x −x−2 =4y,
lg(1 + y) = 2 lg y + lg 2.

Solution: The second equation leads to
lg(1 + y) = lg 2y ⇒ 1+ y =2y ⇒ 2y − y − 1 =0 ⇒ y =1 or −1/2 (deleted since
2
2
2
y> 0). Substitute y =1 into the first equation:
√ 2 √
2
2
2
2 x −x−2 =2 ⇒ x − x − 2 =2 ⇒ x − x − 6 =0 ⇒ (x − 3)(x + 2) = 0 ⇒ x =3 or
x = −2. Hence, (3, 1), (−2, 1) are solutions of the original equation system.
√
2.39 Solve the equation 4x +2x + 7 = 12x +6x − 119.
2
2
√
Solution: Write the equation as 4x +2x + 7 = 3(4x +2x + 7) − 140. Let
2
2
√
4x +2x +7 = t (t ≥ 0), then
2
2
2
t =3t − 140 ⇒ 3t − t − 140 = 0 ⇒ (3t + 20)(t − 7) = 0 ⇒ t = −20/3 (deleted) or
t =7.
√
4x +2x +7 = 7 ⇒ 4x +2x + 7 = 49 ⇒ 2x + x − 21 = 0 ⇒ (x − 3)(2x + 7) =
2
√
Thus 4x +2x +7 = 7 ⇒ 4x +2x + 7 = 49 ⇒ 2x + x − 21 = 0 ⇒ (x − 3)(2x + 7) =
2
2
2
2
2
0 ⇒ x =3
0 ⇒ x =3 or x = −7/2. We can verify that both x =3,x = −7/2 are roots of the
original equation.
2.40 Let S be the sum of reciprocals of two real roots of the equation
(a − 4)x + (2a − 1)x +1 = 0 where a is a real number, find the range of S .
2
2
1−2a
1
1 + x 2 =
Solution: Let x 1 ,x 2 be the two roots, then x =3,x = −7/2 ,x 1 x 2 = a −4 . The quadratic equation
2
2
a −4
2
has real roots, thus a − 4 �=0, Δ = (2a − 1) − 4(a − 4) ≥ 0, thus a �= ±2,a ≤ 17/4. Hence,
2
2
S = 1 + 1 = x 1+x 2 =1 − 2a should satisfy S �= −3,S �=5,S ≥−15/2.
x 1 x 2 x 1x 2
2.41 Let a, b be two real numbers, |a| > 0, and the equation ||x − a|− b| =5 has
three distinct roots, find the value of b .

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