# Need some basic algebra methods....

Will Carpenter

Greenhorn

Posts: 26

posted 11 years ago

- 0

Given two points on a graph (you know, each with a x-coordinate and a y-coordinate) which method could I use to determine the slope of a line between those two points?

Which method could I use to determine the distance between those two points?

Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?

Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?

Which method could I use to determine the distance between those two points?

Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?

Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?

Carlos Failde

Ranch Hand

Posts: 84

posted 11 years ago

If point one has coordinates x1,y1 and point two has coords x2,y2

then the slope is

(y2 - y1) / (x2 - x1)

Pythagoras's Theorem:

distance = sqrt[ (y2-y1)^2 + (x2 - x1)^2 ]

Oh dear how embarrassing that I had to look that one up

One root is x = [ -B + sqrt( B^2 - 4*A*C) ] / (2*A)

The other is x = [ -B - sqrt( B^2 - 4*A*C) ] / (2*A)

Yikes... er ... .... hey what's that thing behind you !? .... *RUNS*

[ March 20, 2004: Message edited by: Carlos Failde ]

- 0

Given two points on a graph (you know, each with a x-coordinate and a y-coordinate) which method could I use to determine the slope of a line between those two points?

If point one has coordinates x1,y1 and point two has coords x2,y2

then the slope is

(y2 - y1) / (x2 - x1)

Which method could I use to determine the distance between those two points?

Pythagoras's Theorem:

distance = sqrt[ (y2-y1)^2 + (x2 - x1)^2 ]

Given three integer coefficients, A, B, C, which method could I use to solve a quadratic function of the form: Ax^2 + Bx + C = 0 ?

Oh dear how embarrassing that I had to look that one up

One root is x = [ -B + sqrt( B^2 - 4*A*C) ] / (2*A)

The other is x = [ -B - sqrt( B^2 - 4*A*C) ] / (2*A)

Given four integer coefficients, A, B, C, D which method could I use to find the derivative of a cube polynomial function: Ax^3 + Bx^2 + Cx + D?

Yikes... er ... .... hey what's that thing behind you !? .... *RUNS*

*(hee hee hee)*

[ March 20, 2004: Message edited by: Carlos Failde ]

chi Lin

Ranch Hand

Posts: 348

Will Carpenter

Greenhorn

Posts: 26

Tim West

Ranch Hand

Posts: 539

posted 11 years ago

- 0

There's a proof in number theory that there's no formula for the zeroes of an order-5 polynomial (ie, p(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f), or any polynomial of greater order.

OK, when I say "formula", I mean a function f(a, b, c, d, e, f) which yields the zeroes to p(x) above.

IIRC there are formulae for the zeroes of polynomials of orders 3 and 4, but they're sufficiently complex that in almost all circumstances an approximation is better.

--Tim

OK, when I say "formula", I mean a function f(a, b, c, d, e, f) which yields the zeroes to p(x) above.

IIRC there are formulae for the zeroes of polynomials of orders 3 and 4, but they're sufficiently complex that in almost all circumstances an approximation is better.

--Tim

Consider Paul's rocket mass heater. |