Quadratic: x^2=2x-3 show a^3=a-6

[From: ] [author: ] [Date: 11-12-27] [Hit: ]

-Man who waits for roast duck to fly into mouth must wait very, very long time.-x^2 - 2x + 3 = 0Well... that leads to two complex roots.......

(1 - (i*√2))^3 = -5 - i√2

(1 + i√2)^3 = -5 + √2

1) a^3=a-6

1a) Minus (in +/-)

-5 - i√2 ?= 1 - i√2 - 6 YES

1b) Plus (in +/-)

-5 + √2 ?= (1 + i√2) - 6 YES

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2) a^2 - 2a^3=9

2a) Minus

(1 - i√2)^2 = -1 - 2i√2

-1 - 2i√2 - 2*(-5 - i√2) ?= 9

-1 + 10 -2-5 - i√2 + -5 + i√2 ?= 9

- i√2 + i√2 = 0 and -1 + 10 = 9

9 ?= 9 YES
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2b) Plus

(1 + i√2)^2 = -1 + 2 i√2

a^2 - 2a^3=9

-1 + 2 i√2 - (2 * (-5 + √2)) ?= 9

10 - 1 + 2 i√2 - 2 i√2 ?= 9 YES

So both of your equations are satisfied by brute strength.

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Man who waits for roast duck to fly into mouth must wait very, very long time.

-

x^2 - 2x + 3 = 0

Well... that leads to two complex roots.... But lets just carry on with your work thus far.

a + b = -2
a b = 3

a = 3/b
3/b + b = -2 (multiply through by b, and b is not zero)
3 + b^2 = -2b
dun dun dun... we are back full circle, so now what?

Well usually we use completing the square for these things since this sort of circular reasoning will always arise. I will just jump straight to the quadratic formula:

x = (-b + sqrt(b^2 - 4ac))/(2/a) and (-b - sqrt(b^2 - 4ac))/(2/a)

Where c = 3 in your deal here (if you've never seen it before). Alas you will find b^2 - 4ac < 0, so you end up with no roots! At least no real roots.
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