Page 67 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

x −2x+4
2.55 Given y =8 2 , find all values of x such that y is an integer.
=
2
x −3x+3
x +1
2
x −2x+4
Solution: y =8 2 = 1+ x+1 . Let λ = , then λx − (3λ + 1)x +3λ − 1 =0
=
2
2
2
x −3x+3 x −3x+3 x − 3x +3
√ √
2 2 5−2 7 5+2 7
( ). The discriminant Δ ≥ 0 ⇒ (3λ + 1) − 4λ(3λ − 1) ≥ 0 ⇒ 3λ − 10λ − 1 ≤ 0 ⇒ ≤ λ ≤
3 3
√ √
2 2 5−2 7 5+2 7 . To make y an integer, λ should be an integer, thus ,
Δ ≥ 0 ⇒ (3λ + 1) − 4λ(3λ − 1) ≥ 0 ⇒ 3λ − 10λ − 1 ≤ 0 ⇒ ≤ λ ≤ λ =0, 1, 2, 3
3 3
√ √
substitute them into ( ) to obtain x = −1, 2+ 2, 2 − 2, 1, 5/2, 2, 4/3.

2
2
2.56 Solve the equation 12 − 12 2 + x − 12 2 = x .
x
x
Solution: Squaring both sides to obtain
√
12 2 12 144 144 4 4 12
2
12 − 2 + x − 2 +2 12x − 2 + 4 − 12 = x ⇔ 2 12 x − x − 12 + 2 =
2
√
x
x 12 x 12 x 2 144 144 4 x − x − 12 + 12
x
2
2
2 + x
4
12 −
4
2
2 )
|x|(x − x − 12 + 24 − x 2 +2 12x − x 2 + x 4 − 12 = x ⇔ 2 12 x 2 =
x
x
2
4
4
2
|x|(x − x − 12 + 24 2 ) . Let x − x − 12 = t and substitute it into the above equality:
x
2
2
2
2 2
2
2 2
2
4 × 12(t + 12 2 ) = x (t + 24 4 + 48t 24 2 = x t + 24 2 + 48t ⇔ x t =0 . Since x =0,
2 ⇔ 48t +
x x x x x
2
2
4
then t =0, then x − x − 12 = 0, then x = 1±7 which should be nonnegative, thus
2
x =4, that is x = ±2. We can easily verify that x = ±2 are roots of the original equation.
2
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