Re: Don thinks 1=0.

John Harshman wrote:

spintronic wrote:


On 4 Oct, 23:02, John Harshman <jharshman.diespam...@xxxxxxxxxxx>
wrote:


spintronic wrote:


On 4 Oct, 21:53, josephus <dogb...@xxxxxxxxxxxxx> wrote:


Barking mad, and so are your posse, of Bob & Bob.

so the discimiate annoys you.

No, you just got it wrong.

the answer was -(sqrt(10)/2 == -B/2

<snip>

Again, I dont think you understand.
X^2 + X + 2,5 = ?.
Has no "real" solutions ending in 0.

That was garbled, but what you probably meant to say is correct. It has
two complex roots, -1/2 + 3i/2 and -1/2 - 3i/2.



Well they are the same, root!

No, they're different roots. + and -, you know.


Just one sec. I'll isolate them. (-1/2 + 3i/2) or (-1/2 - 3i/2).

Nope they are the same root.


This is fascinating. Do you also think that (-1 + 3) is the same as (-1
- 3)? It's usually thought that (A + B) and (A - B) must be different
numbers unless B is zero. What new mathematical insight have you had
that makes this untrue?


but what you probably meant to say is correct. It has
two complex roots, -1/2 + 3i/2 and -1/2 - 3i/2.

.....



It has two complex roots, -1/2 + 3i/2 and -1/2 - 3i/2.

.....


But it does. It has one root: -sqrt(10)/2.


I wonder if you know that those 2 statements are in direct conflict.


No they aren't. Why would you think so?


If a discriminant is 0 there is only one real solution, a double root
where x = -b/2a


It's not just one real solution. It's one solution, period. There is no
complex solution.


According to josephus x = -sqrt(10)/2, a=1, b = -sqrt(10), c = 2.5.

The discriminant = b^2 - 4ac = (-sqrt(10)^2) - (4*1*2.5) = -20.

Wow, would you look at that, the discriminant is NEGATIVE
who would have thought?


I certainly wouldn't. Perhaps you should check your arithmetic again.
What is the square of -sqrt(10)? You seem to think it's -10. No, it's 10.


Now due to the fact that the discriminant is *Actually* NEGATIVE.

That means there are instead 2 Distinct "Non Real" COMPLEX roots.
Called "complex conjugates" of each other.

x = [-b/2a] + [isqrt(4ac - b^2)/2a]
x = [-b/2a] - [isqrt(4ac - b^2)/2a]


If b^2 - 4ac is negative, you do indeed have two complex roots. But if
you will check your math again you will find that it's zero, not
negative. If if were negative, your formula would be right. As it is,
however, it reverts to -b/2a +- 0. One real root.


Now going BACK to my original solution.

X = [(-(2.37955409 * i))^2 = -5.66227767]
0 = [(-(2.37955409 * i))^2 = -5.66227767] + [(10^(1 / 2)) + 2.5 =
5.66227767]


Unless this is a solution to a different equation than the one above,
you have missed a necessary factor of X in the second term. The equation
we were talking about is x^2 + sqrt(10)x + 2.5 =0. The equation you try
to solve here is x^2 + sqrt(10) + 2.5 = 0. Two different equations. The
second equation has a b term of 0. (a=1, b=0, c= sqrt(10) + 2.5).


X = [(2.37955409i)^2 = -5.66227767]
0 = [(2.37955409i)^2 = -5.66227767] + [(10^(1 / 2)) + 2.5 =
5.66227767]

Woops, I was right, and Josephus smells.


Sorry, no.

spin seems to capitalize on typos so he can talk sideways to us. and he does claim to be drunk. the law in the us now says even if you are drunk you can be held accountable.


josephus

--
I go sailing in the Summer and
look at STARS in the Winter.
"Everybody is igernant, jist on differt subjects"
Will Rogers Jr.
"it aint what you know that gets you in trouble
it is what you know that aint so"
Josh Billings.

.

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