Page 80 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

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2.85 Given f(x) =lg(x + 1), solve the equation f(100 − 10 x+1 ) − f(24) = 0.
Solution: The function f(x) =lg(x + 1) has the domain (−∞, +∞), and it is
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decreasing on (−∞, 0) and increasing on (0, +∞). In addition, it is an even function. Hence,

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f(100 − 10 x+1 ) − f(24) = 0 ⇔ f(100 − 10 x+1 )= f(24) ⇔ 100 − 10 x+1 = ±24.
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When 100 − 10 x+1 = 24, we have (10 ) − 10 · 10 − 24 = 0 ⇒ (10 + 2)(10 − 12) = 0 ⇒ 10 = 12 ⇒ x = lg 12
(10 ) − 10 · 10 − 24 = 0 ⇒ (10 + 2)(10 − 12) = 0 ⇒ 10 = 12 ⇒ x = lg 12 since 10 +2 > 0.
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When 100 − 10 x+1 = −24, we have (10 ) − 10 · 10 + 24 = 0 ⇒ (10 − 4)(10 − 6) = 0 ⇒ 10 =4
or 10 =6 ⇒ x = lg 4 or x = lg 6. Therefore, the original equation has three roots:
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x = lg 12,x = lg 4,x = lg 6.
2.86 The equation x + ax + bx + ax +1 = 0 has at least one real root, where
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a, b are real numbers. Find the minimum value of a + b .
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Solution: x =0 is not a root, so we assume x =0 and divide both sides by x to obtain
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(x + ) + a(x + )+ b − 2= 0 (i). (x + ) = x +2+ 1 2 =(x − ) +4 ≥ 4 , thus
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|x + |≥ 2. Let y = x + , then (i) becomes y + ay + b − 2= 0 (|y|≥ 2) (ii). (ii) needs
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x x √ √
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a a −4(b−2) −a± a −4(b−2) |≥ 2 , thus
to have a real root and |y|≥ 2 , then | | + | | ≥|
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a − 4(b − 2) ≥ 4 −|a| . Now we are ready to find the minimum value of a + b . Without
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loss of generality, assume a> 0.

(1) When a ≤ 4, we have a − 4(b − 2) ≥ 4 − a ≥ 0, taking square to obtain 2a ≥ b +2. When
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b +2 ≥ 0, b ≥−2, 4a ≥ b +4b +4, then a + b ≥ (b +4b + 4) + b = (b + ) + .
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Hence, a + b has the minimum value when b = − . When b +2 ≤ 0 , b ≤−2, then
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a + b ≥ b ≥ 4 > .
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(2) When a> 4, we have a + b >a > 16 > .
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As a conclusion from (1)(2), a + b has the minimum value .
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2.87 If a, b are distinct prime numbers, show the x, y -dependent equation
√ √ √
x + y = ab has no positive integer solution.
Proof: We prove the result by contraction. Assume the equation has a positive solution x, y
√ √ √ √ √
such that x + y = ab holds. Taking square to obtain x + y +2 xy = ab , thus xy
is a rational number. xy is a positive integer whose square root is either a positive integer
or a irrational number. Hence, √ xy has to be a positive integer.
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