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Thu 29/11/01 at 23:31
Regular
Posts: 787
I just had to prove that the derivative of the formula for solving quadratic equations was as follows:

========

We know that (x + m) squared = n
where n is greater or equal to 0

has the following solutions:

x = -m +/- sqrt n

So multiply (x + m) squared to give

x sqrd +2mx + m sqrd = n

or

x sqrd + 2mx + (m sqrd - n) = 0

a(x sqrd) + bx + c = 0 can be written as

x sqrd = (b/a)x + (c/a)x = 0

Therefore

2m = b/a and m sqrd - n = c/a

Therefore

m = b/(2a) and n = m sqrd - c/a

Therefore

n = (b/(2a)) sqrd - c/a

n = (b sqrd)/(4(a sqrd)) - c/a

n = ((b sqrd) - (4ac))/(4(a sqrd))

========

Does it look right to you?
Sat 01/12/01 at 18:42
Posts: 0
No its not right.

Look at it again, it may be a typo, but it doesn't look at all right to me.
Sat 01/12/01 at 19:08
Regular
"---SOULJACKER---"
Posts: 5,448
I all honesty FM, I don't understand half the things you've done (mainly due to the inability to use SQAURE signs easily online).

However, there are far better ways to go about proving the quadratic forumla... from what I remember, use "completing the square" on the original quadratic and work from there.

The huge problem with your method is that, from what I can see, it gives the wrong answer! The formula is actually:

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



Oh, having just read what Venom has to say, he has, in fact, come up with the best way to prove that the equation above solves the equation ax^2 + bx =c = 0... just take the one above, and arrange it into the form I nhave just given, and that I proof.

Sonic
Sat 01/12/01 at 19:09
Regular
"---SOULJACKER---"
Posts: 5,448
MISTAKE: should have been:

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

Sonic
Sat 01/12/01 at 19:35
Regular
"---SOULJACKER---"
Posts: 5,448
Ok, I will prove the formula for you:

When I say xSQ I mean x^2, and when I say ROOT(A+B)/3 I mean the root of (a + b) and AFTER THE ROOT IS TAKEN divide by 3.

axSQ + bx +c = 0

(quadratic formula) Now, take out a factor of a:

a(xSQ + (b/a)x + c/a) = 0

Complete the square (if you don't get this, look in any gcse maths text book for "completing the square":

a[(x + (b/2a))SQ - (bSQ)/(4aSQ) + c/a)] = 0

Divide by a:

x + (b/2a)SQ - (bSQ)/(4aSQ) + c/a = 0

(x + (b/2a))SQ = (bSQ)/(4aSQ) - c/a

Now, remember that c/a = 4ac/4aSQ :

(x + (b/2a))SQ = (bSQ)/(4aSQ) - (4ac)/(4aSQ)

(x + (b/2a))SQ = (bSQ -4ac)/(4aSQ)
Take the sq root of the right and left (remember the root can be + or -!):

x + b/2a = +/-[ROOT(bSQ-4ac)/2a]
x = -b/2a +/-[ROOT(bSQ-4ac)/2a]
x = {-b +/- ROOT( bSQ - 4ac)] / 2a


And that's it!

Sonic
Sat 01/12/01 at 19:53
Regular
"---SOULJACKER---"
Posts: 5,448
Well, give me praise!!!!

Sonic
Sat 01/12/01 at 23:21
Regular
"---SOULJACKER---"
Posts: 5,448
No problem Fm...

any time!

Sonic
Sat 01/12/01 at 23:23
Regular
"I like cheese"
Posts: 16,918
Well done SonicRav!!

That was...erm...enthralling...in a sense.
Sat 01/12/01 at 23:26
Regular
"Excommunicated"
Posts: 23,284
FM = 2FM(dumb)
Sat 01/12/01 at 23:28
Regular
"I like cheese"
Posts: 16,918
FM=Galaxy Cake Bar
Sat 01/12/01 at 23:48
Regular
"---SOULJACKER---"
Posts: 5,448
We all know that women take time and money

Women = time X money

Now, time is money.

Women = money X money
Women = money^2

But remember, money is the root of all evil:

Women= Root(evil)^2

Thus,

Women = Evil

Sonic

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