solving quadratics

[Dak]

i had to do this reasently, and i think i may have done it a long winded way.

 

basically, i had a list of the products of a quadratic exuasion for variouse numbers. what i did, was i terated two of them like a simultaniose equasion, ie

 

16a+4b+c=20

4a+2b+c=15

 

then jiggery-pokeried them untill there was only one equasion (times bottom by two; subtract bottom from top)

 

8a-c=-10

 

then i did the same with two different quads, to get another xa + yc = z term.

 

then i treated both the terms as simultaniouse equasions, so that i could work out a term, and then substituted them back into the equsions to work out the other terms.

 

it took ages.

 

is there a quicker (simple) way to do this? bare in mind its been quite a while since iv done advanced maths, so if your replys go along the lines of 'capsised M, funny s 7 arrow 2 wibbly line (3 * (square root of a ternip)) times ((wibbley line) arrow with the word 'series o --> drunk 8' on it [pie (mmm, pie) times by the number of letters in the current day]) divided by [greekletterchi]9, then im not likely to know what your talking about.

[Dave]

I'm a little confused as to the method you're using - or just generally what the problem is. Care to elaborate further?

[Dak]

sorry, i was tired when i posted that.

 

ok:

 

on the left, product of the quadratic eqasion. on the right, the numbers which are input into the equasion (ie, x)

 

735:1

682:2

582:4

535:5

490:6

406:7

367:9

262:12

202:14

175:15

150:16

106:18

87:19

70:20

42:22

 

i had to work out the quadratic equasion used to generate the numbers, and this is what i did:

 

ok, so its quadratic:

 

ax^2 + bx + c = y

 

taking 1 (which gives 735) and 2 (which gives = 682)

 

a + b + c = 735

4a + 2b + c = 682

 

divide second by 2 to get

 

a + b + c = 735

2a + b + 1/2c = 341

 

take second from first to get

 

1/2c -a = 394 <--eqasion A

 

now take 4(which gives 582) and 6(which gives 490) to get

 

16a + 4b + c = 582

36a + 6b + c = 490

 

times top by 3 and bottom by 2 to get

 

48a + 12b + 3c = 1746

72a + 12b + 2c = 980

 

take second from first to get

 

-24a + c = 766 <equasion B

 

put equasions A and B together

 

1/2c -a = 394

c - 24a = 766

 

times top by 2 to get

 

c -2a = 788 <--equasion C

c - 24a = 766

 

take first from second to get

 

-22a = -22

 

times by -1 and divide by 22 to get

 

a = 1

 

using equasion C

 

c -2a = 788

 

becomes

 

c - 2 = 788

 

add 2

 

c = 790

 

now the quadratic for '1', which gives 735

 

a + b + c = 735

 

becomes

 

1 + b + 790 = 735

 

subtract 791

 

b = -56

 

so, the quadratic is

 

x^2 - 56x + 790 = y

 

test:

 

take 22, which gives 42.

 

x^2 - 56x + 790 = y = 42

22^2 - (56*22) + 790 = y = 42

484 -1232 + 790 = y = 42

42 = y = 42

 

right, now all of that took a long time, and essentialy was a prosess of simultaniously solving equasions - but i dont remember it, in the dim resesses of my mind, taking that long when i was at college and could actually remember maths.

 

so my question is, is there a faster/easyer way of doing what i did? the phrases 'bracketing out' and that equasion that involves 'plus-or-minus b times square root of 4ac' spring (vaguely) to mind.

 

cheers

[Dave]

Not that I can see; it seems to be a rather convoluted question.

 

The easiest way to do it that I can see is to just pick three numbers, and then solve the system of 3 simultaneous equations. There are ways of shortening this process, but it involves matrices and row reduction, so I wouldn't suggest it. This is probably one of the quicker ways.

 

FYI: [math]b^2 - 4ac[/math] is the determinant of the quadratic; lets you know how many roots there are and/or if there are any roots of the equation [math]ax^2 + bx + c = 0[/math].

[Dak]

gawd, its amazing how quick you forget this stuff if you dont use it. i used to know all this. roots of the equasion as in the numbers in the bracket? ie

 

[math]3x^2 + 6x + 5 = 0 --> (3x+3)(x+2) = 0[/math] where 3 + 2 would be the roots?

 

umm, and tripple simultaniouse equasions, would that be along the lines of

 

5a + 6b + c = Z

5a + 6b + 12c = Y

5a + 3b + c = X

 

and take 2nd from 1st then 3rd from 1st?

 

sorry this is all basic stuff

[Dave]

That's the idea with the three simultaneous equations, yes. However:

 

[math]3x^2 + 6x + 5 \neq (3x + 3)(x+ 2)[/math]

 

Quickly expanding the brackets will show you this. In fact, that quadratic doesn't have any real roots.

 

But for something like [math]x^2 + 5x + 6[/math], you can factor that to [math](x+3)(x+2)[/math]. Then:

 

[math]x^2 + 5x + 6 = 0 \Rightarrow (x+3)(x+2) = 0 \Rightarrow x = -3, -2[/math].

[Dak]

[math']3x^2 + 6x + 5 \neq (3x + 3)(x+ 2)[/math]

 

DOH!

 

... so the roots are the inverse of the terms in the brackets then?

[Dave]

A quadratic doesn't have any roots until you set it equal to something

 

For example, [math]x^2 + 5x + 6[/math] has factors [math](x+3)[/math] and [math](x+2)[/math]. If [math]x^2 + 5x + 6 = 0[/math] then the equation has roots -2 and -3, but if I said [math]x^2 + 5x + 6 = 2364[/math] then you'd have different roots.

[Dak]

AHA! its comming back now. yes indeedy,

 

If [math]x^2 + 5x + 6 = 0 [/math] then the equation has roots -2 and -3

 

'cos both -2 and -3 will give the answre 0 when plugged into the quadratic.

 

but if I said [math]x^2 + 5x + 6 = 2364[/math] then you'd have different roots.

which i would work out using that plus-or-minus b times sqare root of 4ac over 2a or something formula (sorry, i will learn latex soon)! cool, i can remember (kinda)

 

is it cool to pop back in this forum and ask exessively simple questions everynow and again, i kinda wanna re-learn maths?

[Dave]

You can ask any questions you like in here, within reason And they obviously have to be math-related.

[Dak]

cool, i know 'help me with this' is fine, but wasnt sure if 'randomly teach me stuff' was ok.

 

still, its not as if i have to learn it, just remember it. i aim to get back up to being able to integrate at least.

 

cheers