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"Honest Opinions Please ;-)"

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?

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

MISTAKE: should have been:

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

Sonic

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

No its not right.

Look at it again, it may be a typo, but it doesn't look at all right to me.

honest opinion

HELLLLLLLPPP MATHSSSSSSSSSSSSSS

?? But I thought:

sprd= 2x + 71p

Ummm... Yes... :-)

Yep, looks fine to me.

WAT!!!!! emmmmmmmmm no i dont understand. wwwwwwwwwhhhhhhhhhhhhhhhhhaaaaaaaaaaa!!!!!!

FantasyMeister wrote:
Does it look right to you?

yes

and on the note of workin stuff out
while doing my taxes i accidently proved that God doesnt exist
i would post it but i dont want to