Page 71 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Since m, n are both positive integers, then we only have two possibilities:

m − 1 =1,n − 1= 3 or m − 1 =3,n − 1= 1, which lead to m =2,n =4 or
n
m =4,n =2, thus m + n m = m · n =4 · 2 = 256 .
2
m
n
4
m −n + n −m
2.65 We usually use [x] to represent the integer part of the real number x, here we
define {x} = x − [x] which is the decimal part of the real number x. (1) Find a real number
1
x to satisfy {x} + { } =1. (2) Show that all x satisfying the equation in (1) are not
x
rational numbers.
Solution: (1) Let x = m + α, 1 = n + β (m, n are integers, 0 ≤ α, β ≤ 1).
x
1
{x} + { } =1 ⇔ α + β =1, thus x + 1 = m + α + n + β = m + n +1 is an integer.
x x
2
Let x + 1 = k (k is an integer), that is x − kx +1 = 0 whose roots are
x
√
1 2
x = (k ± k − 4).
2
1
When |k| =2, |x| =1 which does not satisfy the equation {x} + { } =1.
x
√
1 2 1
When |k|≥ 3, x = (k ± k − 4) which satisfies {x} + { } =1.
2 x
2
2
(2) k − 4 is not a perfect square (if it is, then k − 4= h , i.e. k − h =4, but when |k|≥ 3
2
2
2
the difference between two perfect squares is not less than 5), thus x is an irrational number.
2.66 The equation (x − 1)(x − 4) = k has four nonzero real roots, and these roots
2
2
form an arithmetic sequence, find the value of k .
2
Solution: Let y = x , then the equation becomes y − 5y +4 − k =0. Let α, β (0 <α <β )
2
√ √
are roots of y − 5y +4 − k =0, then the original equation has four roots ± α, ± β . They
2
√ √ √ √
form an arithmetic sequence, then β − α = α − (− α), then β =9α . In addition, Vieta’s
9
9
1
formulas imply α + β =5, then we can obtain α = ,β = , thus 4 − k = αβ = , therefore
2 2 4
9 7
k =4 − = .
4 4
2.67 Given a real number d and |d|≤ 1/4, solve the equation
4
3
2
x − 2x + (2d − 1)x + 2(1 − d)x +2d + d =0.
2
Solution: Rewrite the equation as d + (2x − 2x + 2)d + x − 2x − x +2x =0 and treat it a
4
2
3
2
2
2
2
quadratic equation for d , then the quadratic formula implies d = −x − x or d = −x +3x − 2.
Both are quadratic equations for x . Solve them to obtain four roots of the original equation:
√ √ √ √
x = −1+ 1−4d −1− 1−4d 3+ 4d+1 3− 4d+1 . All these roots exist since |d|≤ 1/4.
,
,
,
2 2 2 2
2
2
2.68 Show that the x, y -dependent equation x − y + dx + ey + f =0 represents two
straight lines if and only if d − e − 4f =0.
2
2
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