give me a problem

[SteveCRM]

I am taking algebra 1...we are doing the quadratic formula and graphing etc...someone give me a hard problem to do...

[noble]

this isn't necessarily a hard problem but it's good to be
able to understand and do:

a*x^2 + b*x + c = 0

solve this for x so that x = quadratic formula
another words, prove the quadratic formula

[HarryW]

<-- I don't know what you mean Digital -->

[[Digital-X-Treme]]

what u on about Harry? :P

[noble]

lol, was reading my way down the thread again and
saw harry's post and was like ***?!?!? hehe :P

yeah let steve answer it !!!! :P

[parksie]

Here's another one...use the quadratic formula to prove that all quadratics have two complex roots

[HarryW]

You need to know about complex numbers for that though.

[kedaman]

or you may reinvent the wheel.. Aah, let's say i=sqr(-1)..

[parksie]

(a + bi)^2 = a^2 + b^2

[kedaman]

Hey parksie, put your face back, the forums won't be the same without it

[Sastraxi]

He's on holiday, Kedaman, he can't hear you.. No one can... sob... he's gone...

Oh well, we wont die without Parksie on the forums, but I do like that picture of his face

[noble]

parksie:
don't know if that was a typo but.........

(a+bi)^2 = a^2 - b^2

[da_silvy]

Ok, here's one for anyone to do?

What is the formula to plot a love heart on a Polarmetric graph (i think?)

Calculators may be used..

[TB]

I have tested a lot of functions, but i couldn't display a heart, but a flower: r(a)=sin(a*2)

Here's the proof for the quadratic formula:

ax^2 + bx + c = 0
x^2 + (b/a)x + (c/a) = 0
(x + b/2a)^2 - (b^2/4a^2) + (c/a) = 0
(x + b/2a)^2 = b^2/4a^2 - c/a
x + b/2a = +-sqrt(b^2/4a^2 - c/a)
x + b/2a = +-sqrt(b^2/4a^2 - 4ac/4a^2)
x + b/2a = +-sqrt[(b^2 - 4ac)/4a^2)]
x + b/2a = +-sqrt(b^2 - 4ac) / 2a
    |
    |
    |
   \/

---> x(1,2) = [-b +- sqrt(b^2 - 4ac)] / 2a <---

[da_silvy]

oh well, thanks, i will keep trying and hopefully post the solution here...

[samwise]

lets say you have two equations:
y = x*x + 6*x + 6

and

y = 2*x + 2

at what two / one / none points do they intersect?

where the the quadratic equation intersect the x-axis?
where the the quadratic equation intersect the y-axis?

Samwise Galenorn
[email protected]

e-mail for answer. if you do, make sure to attach which thread this is from.

[jijie]

ok..if you want a problem..I got one here..
how can I read a data in example: notepad, and plot the graph in vb?..
and how we calculate standard deviation/variant(statistical function) in vb, especially in calculating the frequency?..please help me...

[pranavdesai]

if there are two complex nos and if there sum is real as well as their prod is real then prove that they have to be real nos or complex conjugates of each other

[pranavdesai]

that's a good algebra question,what say you?

[KrishnaSantosh]

Hi,

I got a problem and i'm trying to solve it for years together in vain. Actually I invented this problem but can't get through the answer. It would be of immense use to me if i get a satisfactory solution. It would also be useful if the arithmetics involved is simpler to program. Here is the problem. (I'm basically not a mathematics student)

Consider a Series very large number say

9807802301406403657067045034064076023

with unequal intervals,but with the same number of digits.

I want to know if thru any formula i could represent the entire series of these numbers in a shorter way(not as a series but a single number), say like this

92938829938988

or something like that.

if the number 92938829938988 is provided as input, it should be possible for getting the original series of those big numbers.

Any way? Hope i made the question clear.

It would be of immense use to me. Please help.

I don't quite understand your question.

[jp_schwartz]

Take a look at this mega-cool code I wrote for shaping a form like a heart. I´m not sure if I missed some API declarations here. This is the very first time I publish this code. Please enjoy.


in a module:
Public Declare Function SetWindowRgn Lib "user32" (ByVal hwnd As Long, ByVal hrgn As Long, ByVal bRedraw As Boolean) As Long

Public Declare Function CreatePolygonRgn Lib "gdi32" (lpPoint As POINTAPI, ByVal nCount As Long, ByVal nPolyFillMode As Long) As Long

Public Type POINTAPI
x As Long
y As Long
End Type


in a botton:
Const ALTERNATE = 1
Dim sng_I As Single
Dim Pi As Double
Dim vertex(0 To 80) As POINTAPI
Dim int_I As Integer
Dim dbl_Ang As Double
Dim dbl_Radio As Double

Pi = 4 * Atn(1)

int_I = 0
sng_I = 0
Do While sng_I <= 2
dbl_Ang = sng_I * Pi
dbl_Radio = 7 + 6 * Sin(dbl_Ang)
vertex(int_I).x = 20 * (9 + dbl_Radio * Cos(dbl_Ang))
vertex(int_I).y = 20 * (3 + dbl_Radio * Sin(dbl_Ang))

sng_I = sng_I + 0.025
int_I = int_I + 1
Loop
SetWindowRgn hwnd, CreatePolygonRgn(vertex(0), UBound(vertex) + LBound(vertex) + 1, ALTERNATE), True

[jp_schwartz]

That post was for the question of da_silvy: What is the formula to plot a love heart on a Polarmetric graph (i think?)

[jp_schwartz]

If anyone want to solve a really nice trigonometric problem...

Determine the points(x, y) of a generic regular polygon, given the number of sides and the lenght of each side. You could start at (0, 0) for the first point and choose the direction you like.

[kedaman]

a=pi2/s
r=l/(2*sin(a/2))
for n=0 to s-1
  p(n).x=cos(a*n)*r
  p(n).y=sin(a*n)*r
next n