problem

[peterliang]

number 38 district 2 2005

x^3 - x^2 + x - 2

a,b,c are the zeroes

find a^2+b^2+c^2

[AuSmith]

An edit made a new post? idk

[peterliang]

same test

number 14

cot(x - 3pi/4) is

[AuSmith]

If [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula] are roots of a quadratic,

[unparseable or potentially dangerous latex formula]

This is the same formula for a cubic.

[unparseable or potentially dangerous latex formula][unparseable or potentially dangerous latex formula]

guess that means [unparseable or potentially dangerous latex formula] is the answer (nonreal roots).

an identity you can either know or derive when you need to know it:

[unparseable or potentially dangerous latex formula]

Since [unparseable or potentially dangerous latex formula] is the inverse of [unparseable or potentially dangerous latex formula], find [unparseable or potentially dangerous latex formula] and then take the inverse.

[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]

[peterliang]

Ah thanks very much

so for a 3rd degree polynomial, c/a = (ab+bc+ac) ? is that just a rule? like -d/a is the product

[AuSmith]

Yea, [unparseable or potentially dangerous latex formula] is the product of the roots taken 2 at a time (product for quadratics), and [unparseable or potentially dangerous latex formula] is the product of the roots taken 3 at a time (product for cubics).

[AuSmith]

Okay, I'm not sure what caused such a problem, but I fixed my answer in the previous post.

The following identity may help in calculating the sum of the cubes of the roots (again the formula works for any polynomial). These are called the Newton-Girard formulas.

[unparseable or potentially dangerous latex formula]