F.5 Maths

x^5 - 29x^3 + 100x = 0
x(x^4 - 29x^2 + 100) = 0
x[(x^2)^2 - 29(x^2) + 100] = 0
x(x^2 - 4)(x^2 - 25) = 0
x(x^2 - 2^2)(x^2 - 5^2) = 0
x(x + 2)(x - 2)(x + 5)(x - 5) = 0
x = 0 or x = -2 or x = 2or x = -5 or x = 5


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8x^5 - 7x^3 - x = 0
x(8x^4 - 7x^2 - 1) = 0
x[8(x^2)^2 - 7(x^2) - 1] = 0
x(x^2- 1)(8x^2 + 1) = 0
x(x + 1)(x - 1)(8x^2 + 1) = 0
x = 0 or x = -1 or x = 1or x = (√2)i/4 or x= -(√2)i/4
(Only the first three roots are real roots.)


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5^2x - 3(5^x) - 4 = 0
(5^x)^2 - 3(5^x) - 4 = 0
(5^x - 4)(5^x + 1) = 0
5^x = 4 or 5^x = -1 (rejected)
log(5^x) = log(4)
xlog(5) = log(4)
x = log(4)/log(5)

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Q 1 : x^5-29x^3+100x=0

A 1 :

x^5-29 x^3+100x = 0
Divide both sides by x, which gives the solution:
x = 0 or x^4-29x^2+100 = 0
Substitute u = x^2 into the left hand side:
x = 0 or u^2-29u+100 = 0
Subtract 100 from both sides:
x = 0 or u^2-29u = -100
Add 841/4 to both sides:
x = 0 or u^2-29 u+841/4 = 441/4
Factor the left hand side:
x = 0 or (u-29/2)^2 = 441/4
Take the square root of both sides:
x = 0 or |(u-29/2)| = 21/2
Eliminate the absolute value:
x = 0 or u-29/2 = -21/2 or u-29/2 = 21/2
Add 29/2 to both sides:
x = 0 or u = 4 or u-29/2 = 21/2
Substitute back for u = x^2:
x = 0 or x^2 = 4 or u-29/2 = 21/2
Take the square root of both sides:
x = 0 or x = -2 or x = 2 or u-29/2 = 21/2
Add 29/2 to both sides:
x = 0 or x = -2 or x = 2 or u = 25
Substitute back for u = x^2:
x = 0 or x = -2 or x = 2 or x^2 = 25
Take the square root of both sides:
x = 0 or x = -2 or x = 2 or x = -5 or x = 5

Answer : x = -5 or -2 or 0 or 2 or 5

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由於字數限制，Q2 及 Q3 將到意見欄回答，不便之處，敬請原諒。

2011-07-21 15:55:52 補充：
Q 2 : 8x^5-7x^3-x=0

A 2 :

8x^5-7x^3-x = 0
Divide both sides by x, which gives the solution:
x = 0 or 8x^4-7x^2-1 = 0
Substitute u = x^2 into the left hand side:
x = 0 or 8u^2-7u-1 = 0

2011-07-21 15:56:01 補充：
Solve the quadratic equation by completing the square:

Divide both sides by 8:
x = 0 or u^2-(7u)/8-1/8 = 0
Add 1/8 to both sides:
x = 0 or u^2-(7 u)/8 = 1/8
Add 49/256 to both sides:
x = 0 or u^2-(7 u)/8+49/256 = 81/256
Factor the left hand side:
x = 0 or (u-7/16)^2 = 81/256

2011-07-21 15:56:27 補充：
Take the square root of both sides:
x = 0 or |(u-7/16) = 9/16
Eliminate the absolute value:
x = 0 or u-7/16 = -9/16 or u-7/16 = 9/16
Add 7/16 to both sides:
x = 0 or u = -1/8 or u-7/16 = 9/16
Substitute back for u = x^2:
x = 0 or x^2 = -1/8 or u-7/16 = 9/16

2011-07-21 15:56:50 補充：
Take the square root of both sides:
x = 0 or x = -i/(2√(2)) or x = i/(2√(2)) or u-7/16 = 9/16
Add 7/16 to both sides:
x = 0 or x = -i/(2√(2)) or x = i/(2√(2)) or u = 1
Substitute back for u = x^2:
x = 0 or x = -i/(2√(2)) or x = i/(2√(2)) or x^2 = 1

2011-07-21 15:56:57 補充：
Take the square root of both sides:
x = 0 or x = -i/(2√(2)) or x = i/(2√(2)) or x = -1 or x = 1

Answer : x = -1 or x = 0 , x = 1 or x = -i/(2√(2)) or x = i/(2√(2)) .

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2011-07-21 15:57:04 補充：
Q 3 : 5^2x-3(5^x)-4=0

A 3 :

5^2x-3(5^x)-4=0
Substitute u = e^(x^2) into the left hand side:
u^2-3u-4 = 0
Add 4 to both sides:
u^2-3u = 4

2011-07-21 15:57:11 補充：
Add 9/4 to both sides:
u^2-3u+9/4 = 25/4
Factor the left hand side:
(u-3/2)^2 = 25/4
Take the square root of both sides:
abs(u-3/2) = 5/2
Eliminate the absolute value:
u-3/2 = -5/2 or u-3/2 = 5/2

2011-07-21 15:57:20 補充：
Add 3/2 to both sides:
u = -1 or u-3/2 = 5/2
Substitute back for u = 5^x:
5^x = -1 or u-3/2 = 5/2
Take the logarithm to the base 5 of both sides:
x = (iπ)/(log(5)) or u-3/2 = 5/2

2011-07-21 15:57:23 補充：
Add 3/2 to both sides:
x = (iπ)/(log(5)) or u = 4
Substitute back for u = 5^x:
x = (iπ)/(log(5)) or 5^x = 4
Take the logarithm to the base 5 of both sides:
x = (iπ)/(log(5)) or x = (log(4))/(log(5))

Answer : x = (log(4))/(log(5)) or (iπ)/(log(5))

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