Page 86 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Solution:
2
2
2
y
0
y
3 ·9 −10y ·81 −11 =1 ⇔ 3 ·3 −20y ·3 −44 =3 ⇒ y −20y−44 = 0 ⇒ (y+2)(y−22) =
2
2
y
y
2
0
3 ·9
0 ⇒ y = −2 −10y ·81 −11 =1 ⇔ 3 ·3 −20y ·3 −44 =3 ⇒ y −20y−44 = 0 ⇒ (y+2)(y−22) =
0 ⇒ y = −2 or y = 22. We only need positive integer roots, so y = 22. The coefficients
n
of the last three terms are C n n−2 + C n n−1 + C = 22, then
n
n(n−1)
2 2 1 1 n(n−1) 2 2
+n = 21 ⇒ n +n−42 = 0 ⇒ (n+7)(n−6) =0 ⇒ n =6
C +C +1 = 2 ⇒ 2 2 +n = 21 ⇒ n +n−42 = 0 ⇒ (n+7)(n−6) =0 ⇒ n =6 since
C +C +1 = 2 ⇒
n
n
n
n
n +7 > 0.
√ √ √ √ √ √ √ √

m
−22
3 3 m =0 − + + 2 2m ⇔m ⇔ 3 3 2m = 100 2 2m ⇒m ⇒ 2 2m = 200 ⇒ m = 20000m = 200 ⇒ m = 20000.
=0.1.1
2m = 100 + +
2 2 2 2
Since n =6, the middle term of the expansion is
3
) = 20x
) =
T 4 = C (x lg x 3 6×5×4 (x lg x 3 3 lg x . According to the condition of the problem, we
6
3×2×1
3 lg x x
3 lg 3 lg 3 lg 2 2
3 lg x x
3 lg x x
= lg 1000 ⇒ 3(lg x) =3 ⇒ lg x
= lg 1000 ⇒ 3(lg x) =3 ⇒ lg x = =
20x lg
= 1000 ⇒ lg x x lg x
= 20000 ⇒ x x lg x
= 20000 ⇒3
= 1000 ⇒ lg 3
2
have 20xx = 20000 ⇒ x = 1000 ⇒ lg x = lg 1000 ⇒ 3(lg x) =3 ⇒ lg x =
3
20x
±1 ⇒ x = 10
±1 ⇒ x = 10
±1 ⇒ x = 10 or x =1/10.
2.95 Let p be an odd prime number, find all positive integer roots of the equation
x = y(y + p).
2
2
Solution: x = y(y + p) ⇔ (x + y)(x − y)= py . Since p is a prime number, we have
p|x − y or p|x + y . If p|x − y , then x − y ≥ p (note that x> y ), thus we should have
x + y ≤ y , impossible. Thus p|x + y. Let x + y = pn (i), where n is a positive integer,
then the original equation becomes n(x − y)= y , thus n|y and x = n+1 y . Hence,
n
2
2
x + y = 2n+1 y (ii). (i)&(ii) lead to (2n + 1)y = n p . Since (n , 2n + 1) = 1, we have
n
2
n |y . (2n + 1) y 2 = p and p is a prime, thus y 2 =1, then p =2n +1 ⇒ n = p−1 , then
n n 2
2
2
y = n =( p−1 2 n+1 y = n+1 2 p−1 · p+1 = p −1 . Therefore, the original
) , x =
n = n(n + 1) =
2 n n 2 2 2
2
) .
equation has only one positive integer root: x = p −1 ,y =( p−1 2
2 2
2.96 Consider the real coefficient equations
2
ax + bx 1 + c = x 2
1
2
ax + bx 2 + c = x 3
2
. . .
ax 2 n−1 + bx n−1 + c = x n
2
ax + bx n + c = x 1
n
where a =0, show that when Δ=(b − 1) − 4ac =0, this equation system has a unique solution.
2
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