Page 77 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

2.80 Solve the equation x + log 3 = log (4 − 5 · 9 ).
x
x
9
9
x
x
x
x
Solution: The equation is equivalent to log 9 + log 3 = log (4 − 5 · 9 ) ⇒ log 3 3x = log (4 − 5 · 9 ) ⇒ 3 3x =4 − 5 · 3 2x
9
9
9
9
9
x
x
x
x
x
log 9 + log 3 = log (4 − 5 · 9 ) ⇒ log 3 3x = log (4 − 5 · 9 ) ⇒ 3 3x =4 − 5 · 3 2x . Let 3 = y , then the equation becomes
9
9
9
9
9

y +5y − 4 =0 ⇒ y + y +4y − 4 =0 ⇒ (y + 1)(y +4y − 4) = 0 ⇒ y = −1y +5y − 4 =0 ⇒ y + y +4y − 4 =0 ⇒ (y + 1)(y +4y − 4) = 0 ⇒ y = −1 or
2
2
2
3
3
3
3
2
2
2
2
2
√ √ √
x
y = −2(1 + 2) or y = 2( 2 − 1). Since y =3 > 0, then y = −1,y = −2(1 + 2) are
√ √ √ √ √ √
x x
incorrect. Hence, y = 2( 2 − 1) ⇒ 3 = 2( 2 − 1) ⇒ x = log 2( 2 − 1)y = 2( 2 − 1) ⇒ 3 = 2( 2 − 1) ⇒ x = log 2( 2 − 1), which is the
3 3
only root.
2.81 Solve the system of equations
4x 2
= y,
1+4x 2
4y 2
= z,
1+4y 2
4z 2
= x.
1+4z 2
Solution: Obviously x ≥ 0,y ≥ 0,z ≥ 0. The first equation together with
2
1+4x = (1 − 2x) +4x ≥ 4x leads to y =8 4x 2 2 ≤ 4x 2 = x . Similarly, the second and
2
=
4x
1+4x
the third equations lead to z ≤ y, x ≤ z . Hence, x = y = z , then
2 2
22
4x4x 2 = x ⇒ 4x − 4x + x =0 ⇒ x(2x − 1) =0 ⇒ x =0= x ⇒ 4x − 4x + x =0 ⇒ x(2x − 1) =0 ⇒ x =0 or x =1/2. Therefore,
33
22

1+4x1+4x 2
(0, 0, 0), (1/2, 1/2, 1/2) are the solutions.
2
2
3
2
3
2.82 Find all distinct real roots of the equation (x − 3x + x − 2)(x − x − 4x +7)+6x − 15x + 18 = 0
2
3
(x − 3x + x − 2)(x − x − 4x +7)+6x − 15x + 18 = 0.
3
2
2
5
9
Solution: Let A = x − 2x + x + ,B = x − x + , then the equation becomes
3
3
2
5
2
2 2 2 2
2
2
(A − B)(A + B) +6B − 9 =0 ⇒ A − (B − 3) =0 ⇒ (A + B − 3)(A − B + 3) =
2
2
(A − B)(A + B) +6B − 9 =0 ⇒ A − (B − 3) =0 ⇒ (A + B − 3)(A − B + 3) =
0 ⇒ A + B − 3= 0
0 ⇒ A + B − 3= 0 or A − B +3 = 0.
If A + B − 3= 0, then x − x − 4x +4 = 0 ⇒ (x − 1)(x − 2)(x + 2) = 0 ⇒ x =1 or
2
3
x = ±2.
If A − B +3 = 0, then x − 3x + x +1 = 0 ⇒ (x − 1)(x − 2x − 1) = 0 ⇒ x =1 or
2
3
2
√
x =1 ± 2.
√
As a conclusion, the equation has four distinct real roots: x =1,x = ±2,x =1 ± 2.
√ √ √
2.83 If a, b, c are real numbers, ac < 0, 2a + 3b + 5c =0, show the quadratic
3
2
equation ax + bx + c =0 has a root within the interval ( , 1).
4
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